Given: Current, I = 0.5 A.

Time, t = 10 minutes. Convert time to seconds: $t = 10 \times 60 \text{ s} = 600 \text{ s}$.

The amount of electric charge (Q) is given by $Q = I \times t$:

$Q = (0.5 \text{ A}) \times (600 \text{ s})$

$Q = 300 \text{ C}$.

The amount of electric charge that flows through the circuit is 300 coulombs.

An electric circuit is a continuous and closed path for electric current to flow. It typically consists of a source of electric energy (like a cell or battery), connecting wires, and one or more electrical components (like a bulb, resistor, switch).

The SI unit of electric current is the **ampere (A)**. One ampere is defined as the flow of one coulomb of electric charge per second ($1 \text{ A} = 1 \text{ C/s}$).

The charge of one electron is approximately $e = 1.6 \times 10^{-19}$ C.

To find the number of electrons (n) in one coulomb (Q = 1 C) of charge, we use the relation $Q = n \times e$:

$1 \text{ C} = n \times (1.6 \times 10^{-19} \text{ C})$

$n = \frac{1 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = \frac{1}{1.6 \times 10^{-19}} = \frac{10^{19}}{1.6}$

$n = \frac{10}{1.6} \times 10^{18} = \frac{100}{16} \times 10^{18} = 6.25 \times 10^{18}$.

Approximately $6.25 \times 10^{18}$ electrons constitute one coulomb of charge. (The textbook rounds this to $6 \times 10^{18}$).

Given: Charge, Q = 2 C.

Potential difference, V = 12 V.

Work done, W = ?

Using the formula $V = W/Q$, we can find W = V $\times$ Q:

$W = (12 \text{ V}) \times (2 \text{ C})$

$W = 24 \text{ J}$.

24 joules of work is done in moving the charge.

A device that helps to maintain a potential difference across a conductor is an **electric cell** or a **battery** (a combination of cells).

Saying that the potential difference between two points is 1 V means that 1 joule of work is done to move 1 coulomb of electric charge from one point to the other against the electric field.

A 6 V battery provides a potential difference of 6 volts across its terminals. The potential difference is defined as the work done (energy given) per unit charge ($V = W/Q$).

Here, V = 6 V and Q = 1 coulomb.

Energy given (Work done), W = V $\times$ Q = (6 V) $\times$ (1 C) = 6 J.

So, 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.

A battery of three cells of 2V each means a total potential difference of $3 \times 2\text{V} = 6\text{V}$. The resistors (5$\Omega$, 8$\Omega$, 12$\Omega$) and the key are connected in series with the battery.

To measure the current through the resistors, the ammeter must be connected in series anywhere in the circuit. To measure the potential difference across the 12$\Omega$ resistor, the voltmeter must be connected in parallel across its ends.

Calculation of readings:

The resistors are connected in series, so the total resistance (R$_s$) is the sum of individual resistances:

R$_s = 5\Omega + 8\Omega + 12\Omega = 25\Omega$.

The total voltage (V) of the battery is $3 \times 2\text{V} = 6\text{V}$.

According to Ohm's Law ($V = IR$), the total current (I) flowing through the circuit is:

$I = V/R_s = 6\text{V} / 25\Omega = 0.24\text{ A}$.

Since the ammeter is in series with the resistors, it measures the total current. Ammeter reading = **0.24 A**.

The voltmeter measures the potential difference across the 12$\Omega$ resistor. Using Ohm's Law for the 12$\Omega$ resistor, $V_{12\Omega} = I \times R_{12\Omega}$:

$V_{12\Omega} = (0.24\text{ A}) \times (12\Omega) = 2.88\text{ V}$.

Voltmeter reading = **2.88 V**.

(a) Given: Voltage, V = 220 V; Resistance of bulb filament, R = 1200 $\Omega$.

Using Ohm's Law, $I = V/R$:

$I = 220 \text{ V} / 1200 \Omega = 22/120 \text{ A} = 11/60 \text{ A} \approx 0.183\text{ A}$.

The electric bulb will draw approximately 0.18 A of current.

(b) Given: Voltage, V = 220 V; Resistance of heater coil, R = 100 $\Omega$.

Using Ohm's Law, $I = V/R$:

$I = 220 \text{ V} / 100 \Omega = 2.2\text{ A}$.

The electric heater coil will draw 2.2 A of current.

First, we need to find the resistance of the heater coil using the initial conditions.

Given: Initial Potential difference, V$_1$ = 60 V; Initial Current, I$_1$ = 4 A.

According to Ohm's Law, the resistance (R) of the heater is constant (assuming temperature remains relatively unchanged):

$R = V_1 / I_1 = 60 \text{ V} / 4 \text{ A} = 15 \Omega$.

Now, we find the current when the potential difference is increased.

Given: New Potential difference, V$_2$ = 120 V; Resistance, R = 15 $\Omega$.

Using Ohm's Law, the new current (I$_2$) is:

$I_2 = V_2 / R = 120 \text{ V} / 15 \Omega = 8 \text{ A}$.

The heater will draw a current of 8 A if the potential difference is increased to 120 V.

Given:

Resistance, R = 26 $\Omega$.

Length, l = 1 m.

Diameter, d = 0.3 mm. Convert to metres: $d = 0.3 \times 10^{-3}$ m = $3 \times 10^{-4}$ m.

Area of cross-section, A = $\pi (d/2)^2 = \pi (3 \times 10^{-4} / 2)^2 = \pi (1.5 \times 10^{-4})^2 = \pi \times 2.25 \times 10^{-8} \text{ m}^2$.

Using the formula $R = \rho \frac{l}{A}$, we can find resistivity $\rho = \frac{R A}{l}$:

$\rho = \frac{(26 \Omega) \times (\pi \times 2.25 \times 10^{-8} \text{ m}^2)}{1 \text{ m}}$

$\rho = 26 \times \pi \times 2.25 \times 10^{-8} \Omega \text{ m}$

$\rho \approx 26 \times 3.14159 \times 2.25 \times 10^{-8} \Omega \text{ m}$

$\rho \approx 183.7 \times 10^{-8} \Omega \text{ m} = 1.837 \times 10^{-6} \Omega \text{ m}$.

The resistivity is approximately $1.84 \times 10^{-6} \Omega$ m.

Predicting the material: Referring to Table 11.2 (in the textbook), the resistivity of **Manganese** is $1.84 \times 10^{-6} \Omega$ m. Thus, the material of the wire is likely Manganese.

Let the resistivity of the material be $\rho$. For the first wire:

$R_1 = \rho \frac{l}{A} = 4 \Omega$.

For the second wire, the length is $l_2 = l/2$ and the area of cross-section is $A_2 = 2A$. The material is the same, so the resistivity $\rho$ is also the same.

The resistance of the second wire ($R_2$) is:

$R_2 = \rho \frac{l_2}{A_2} = \rho \frac{l/2}{2A} = \rho \frac{l}{4A}$.

We can write $R_2$ in terms of $R_1$ by rearranging the expression for $R_1$: $\rho \frac{l}{A} = R_1 = 4 \Omega$.

$R_2 = \frac{1}{4} \left( \rho \frac{l}{A} \right) = \frac{1}{4} R_1$.

$R_2 = \frac{1}{4} \times 4 \Omega = 1 \Omega$.

The resistance of the second wire would be 1 $\Omega$.

The resistance of a conductor depends on the following factors:

1. **Length (l):** Resistance is directly proportional to the length.
2. **Area of cross-section (A):** Resistance is inversely proportional to the area of cross-section.
3. **Nature of the material:** Different materials have different resistivities.
4. **Temperature:** Resistance of a conductor varies with temperature.

Current will flow more easily through a **thick wire** of the same material and length, when connected to the same source. This is because the resistance of a conductor is inversely proportional to its area of cross-section ($R \propto 1/A$). A thick wire has a larger area of cross-section compared to a thin wire. Therefore, a thick wire has lower resistance, allowing current to flow more easily according to Ohm's Law ($I = V/R$).

According to Ohm's Law, $V = IR$, or $I = V/R$. The resistance (R) is given as constant. The current (I) is directly proportional to the potential difference (V).

If the potential difference (V) decreases to half of its former value ($V' = V/2$), while the resistance remains constant ($R' = R$), the new current ($I'$) will be:

$I' = V'/R' = (V/2) / R = \frac{1}{2} (V/R)$.

Since $I = V/R$, the new current $I' = \frac{1}{2} I$.

The current through the component will **decrease to half of its former value**.

Coils of electric toasters and electric irons are made of an alloy (like Nichrome) rather than a pure metal for two main reasons:

1. **Higher Resistivity:** Alloys generally have a higher electrical resistivity than their constituent pure metals. This means that for a given length and thickness, an alloy wire will offer higher resistance than a pure metal wire. According to Joule's law of heating ($H = I^2Rt$), higher resistance produces more heat for the same current, leading to efficient heating.
2. **Higher Melting Point and Resistance to Oxidation:** Alloys, particularly those used in heating elements, have high melting points and do not oxidise (burn) readily at high temperatures. This prevents the heating element from melting or degrading quickly when heated to glowing temperatures, ensuring a longer lifespan for the appliance.

A better conductor has lower resistivity. From Table 11.2 (in the textbook):

• Resistivity of Iron = $10.0 \times 10^{-8} \Omega$ m
• Resistivity of Mercury = $94.0 \times 10^{-8} \Omega$ m

(a) Comparing the resistivity values, Iron ($10.0 \times 10^{-8} \Omega$ m) has lower resistivity than Mercury ($94.0 \times 10^{-8} \Omega$ m). Therefore, **Iron** is a better conductor than mercury.

(b) Looking at the resistivity values for all materials listed as conductors in Table 11.2, **Silver** has the lowest resistivity ($1.60 \times 10^{-8} \Omega$ m). Therefore, Silver is the best conductor among the materials listed.

Given:

Resistance of the electric lamp, R$_1$ = 20 $\Omega$.

Resistance of the conductor, R$_2$ = 4 $\Omega$.

Battery voltage, V = 6 V.

The lamp and the conductor are connected in series.

(a) Total resistance of the circuit (equivalent resistance of series combination, R$_s$):

$R_s = R_1 + R_2 = 20\Omega + 4\Omega = 24\Omega$.

The total resistance of the circuit is 24 $\Omega$.

(b) Current through the circuit (total current, I):

Using Ohm's Law, $I = V/R_s$:

$I = 6\text{V} / 24\Omega = 0.25\text{ A}$.

The current through the circuit is 0.25 A.

(c) Potential difference across the electric lamp (V$_1$) and the conductor (V$_2$). The current through both is the total current I = 0.25 A.

Using Ohm's Law for the lamp: $V_1 = I \times R_1 = (0.25\text{ A}) \times (20\Omega) = 5\text{ V}$.

Using Ohm's Law for the conductor: $V_2 = I \times R_2 = (0.25\text{ A}) \times (4\Omega) = 1\text{ V}$.

The potential difference across the electric lamp is 5 V, and across the conductor is 1 V. Note that V$_1 + V_2 = 5\text{V} + 1\text{V} = 6\text{V}$, which equals the total battery voltage V.

Given:

Resistances: R$_1$ = 5 $\Omega$, R$_2$ = 10 $\Omega$, R$_3$ = 30 $\Omega$.

Battery voltage, V = 12 V.

The resistors are connected in parallel, so the potential difference across each resistor is V = 12 V.

(a) Current through each resistor:

Using Ohm's Law ($I = V/R$) for each resistor:

Current through R$_1$, $I_1 = V/R_1 = 12\text{V} / 5\Omega = 2.4\text{ A}$.

Current through R$_2$, $I_2 = V/R_2 = 12\text{V} / 10\Omega = 1.2\text{ A}$.

Current through R$_3$, $I_3 = V/R_3 = 12\text{V} / 30\Omega = 0.4\text{ A}$.

(b) Total current in the circuit (I):

$I = I_1 + I_2 + I_3 = 2.4\text{A} + 1.2\text{A} + 0.4\text{A} = 4.0\text{ A}$.

The total current in the circuit is 4.0 A.

(c) Total circuit resistance (equivalent resistance of parallel combination, R$_p$):

Using the formula $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$:

$\frac{1}{R_p} = \frac{1}{5\Omega} + \frac{1}{10\Omega} + \frac{1}{30\Omega}$

Find a common denominator (30):

$\frac{1}{R_p} = \frac{6}{30} + \frac{3}{30} + \frac{1}{30} = \frac{6+3+1}{30} = \frac{10}{30} = \frac{1}{3}$.

$R_p = 3 \Omega$.

The total circuit resistance is 3 $\Omega$. (Note that this is less than the smallest individual resistance, 5 $\Omega$).

The circuit consists of two parallel combinations connected in series. R$_1$ and R$_2$ are in parallel. R$_3$, R$_4$, and R$_5$ are in parallel. These two parallel groups are in series with each other.

Given resistances: R$_1$ = 10 $\Omega$, R$_2$ = 40 $\Omega$, R$_3$ = 30 $\Omega$, R$_4$ = 20 $\Omega$, R$_5$ = 60 $\Omega$.

Battery voltage, V = 12 V.

(a) Total resistance in the circuit:

First, find the equivalent resistance of the parallel combination of R$_1$ and R$_2$ (let's call it R'$_{p1}$):

$\frac{1}{R'_{p1}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10\Omega} + \frac{1}{40\Omega} = \frac{4}{40} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8}$.

$R'_{p1} = 8 \Omega$.

Next, find the equivalent resistance of the parallel combination of R$_3$, R$_4$, and R$_5$ (let's call it R'$_{p2}$):

$\frac{1}{R'_{p2}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{30\Omega} + \frac{1}{20\Omega} + \frac{1}{60\Omega}$.

Find a common denominator (60):

$\frac{1}{R'_{p2}} = \frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{2+3+1}{60} = \frac{6}{60} = \frac{1}{10}$.

$R'_{p2} = 10 \Omega$.

These two equivalent resistances, R'$_{p1}$ and R'$_{p2}$, are connected in series. The total resistance of the circuit (R$_s$) is the sum of these equivalent resistances:

$R_s = R'_{p1} + R'_{p2} = 8\Omega + 10\Omega = 18\Omega$.

The total resistance in the circuit is 18 $\Omega$.

(b) Total current flowing in the circuit (I):

Using Ohm's Law ($I = V/R_s$) for the entire circuit:

$I = 12\text{V} / 18\Omega = 2/3 \text{ A} \approx 0.67\text{ A}$.

The total current flowing in the circuit is $2/3$ A (approximately 0.67 A).

When resistors are connected in parallel, the reciprocal of the equivalent resistance (R$_p$) is the sum of the reciprocals of individual resistances: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ...$. The equivalent resistance in parallel is always less than the smallest individual resistance.

(a) Two resistors in parallel: R$_1 = 1\Omega$, R$_2 = 10^6 \Omega$.

$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^6} = 1 + 0.000001 = 1.000001$.

$R_p = \frac{1}{1.000001} \approx 0.999999 \Omega$.

The equivalent resistance is approximately **1 $\Omega$**. It is slightly less than the smallest resistance (1 $\Omega$).

(b) Three resistors in parallel: R$_1 = 1\Omega$, R$_2 = 10^3 \Omega$, R$_3 = 10^6 \Omega$.

$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} = 1 + 0.001 + 0.000001 = 1.001001$.

$R_p = \frac{1}{1.001001} \approx 0.998999 \Omega$.

The equivalent resistance is approximately **0.999 $\Omega$**. It is slightly less than the smallest resistance (1 $\Omega$).

The lamp, toaster, and water filter are connected in parallel to a 220 V source. This means the potential difference across each appliance is 220 V.

Given resistances: R$_{lamp}$ = 100 $\Omega$, R$_{toaster}$ = 50 $\Omega$, R$_{filter}$ = 500 $\Omega$.

Source voltage, V = 220 V.

First, calculate the current through each appliance using Ohm's Law ($I = V/R$):

Current through lamp, $I_{lamp} = V/R_{lamp} = 220\text{V} / 100\Omega = 2.2\text{ A}$.

Current through toaster, $I_{toaster} = V/R_{toaster} = 220\text{V} / 50\Omega = 4.4\text{ A}$.

Current through filter, $I_{filter} = V/R_{filter} = 220\text{V} / 500\Omega = 0.44\text{ A}$.

The total current drawn by the three appliances when connected in parallel is the sum of individual currents:

$I_{total} = I_{lamp} + I_{toaster} + I_{filter} = 2.2\text{A} + 4.4\text{A} + 0.44\text{A} = 7.04\text{ A}$.

An electric iron is connected to the same 220 V source and takes the same amount of current, $I_{iron} = I_{total} = 7.04\text{ A}$.

The resistance of the electric iron ($R_{iron}$) can be calculated using Ohm's Law ($V = I \times R$), where V = 220 V and I = 7.04 A:

$R_{iron} = V / I_{iron} = 220\text{V} / 7.04\text{ A}$.

$R_{iron} = \frac{220}{7.04} = \frac{22000}{704} = \frac{2750}{88} = \frac{1375}{44} = \frac{125}{4} = 31.25 \Omega$.

The resistance of the electric iron is 31.25 $\Omega$, and the current through it is 7.04 A.

Advantages of connecting electrical devices in parallel:

1. **Independent Operation:** Each device works independently of the others. If one device fails or is switched off, the others continue to function. In a series circuit, if one component fails, the entire circuit breaks.
2. **Same Voltage:** All devices receive the full voltage of the battery (or power source), which is usually the required operating voltage for each device. In a series circuit, the voltage gets divided among the devices, meaning each device receives a smaller voltage than the battery voltage (unless they have vastly different resistances).
3. **Different Current Requirements:** Devices with different resistances can operate properly in parallel as they draw current according to their resistance and the applied voltage ($I = V/R$). In series, all devices would need to handle the same current, which might not be suitable for their design.
4. **Lower Total Resistance:** Connecting resistors in parallel reduces the total equivalent resistance of the circuit ($\frac{1}{R_p} = \sum \frac{1}{R_i}$). This allows for a larger total current from the source compared to a series connection of the same devices, which is often desirable for powering multiple appliances.

We have resistors R$_1 = 2\Omega$, R$_2 = 3\Omega$, R$_3 = 6\Omega$. We need to find combinations to get total resistances of 4$\Omega$ and 1$\Omega$.

Possible combinations include series, parallel, or combinations of series and parallel.

Series combination: $R_s = 2+3+6 = 11\Omega$ (not 4$\Omega$ or 1$\Omega$).

Parallel combination: $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1$. $R_p = 1\Omega$.

Combinations of series and parallel:

(a) Target resistance = 4 $\Omega$. We can try connecting some resistors in parallel and then in series with the third one.

• Connect 2$\Omega$ and 3$\Omega$ in parallel, then in series with 6$\Omega$: $\frac{1}{R_{p1}} = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6}$. $R_{p1} = \frac{6}{5} = 1.2\Omega$. Total resistance = $R_{p1} + 6\Omega = 1.2 + 6 = 7.2\Omega$ (not 4$\Omega$).
• Connect 2$\Omega$ and 6$\Omega$ in parallel, then in series with 3$\Omega$: $\frac{1}{R_{p2}} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3}$. $R_{p2} = \frac{3}{2} = 1.5\Omega$. Total resistance = $R_{p2} + 3\Omega = 1.5 + 3 = 4.5\Omega$ (not 4$\Omega$).
• Connect 3$\Omega$ and 6$\Omega$ in parallel, then in series with 2$\Omega$: $\frac{1}{R_{p3}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$. $R_{p3} = 2\Omega$. Total resistance = $R_{p3} + 2\Omega = 2 + 2 = 4\Omega$. **This works!**

To get a total resistance of 4 $\Omega$, connect the 3$\Omega$ and 6$\Omega$ resistors in parallel, and then connect this parallel combination in series with the 2$\Omega$ resistor.

(b) Target resistance = 1 $\Omega$. We calculated the parallel combination earlier:

$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$. $R_p = 1\Omega$. **This works!**

To get a total resistance of 1 $\Omega$, connect all three resistors (2$\Omega$, 3$\Omega$, and 6$\Omega$) in parallel.

We have four resistors: R$_1 = 4\Omega$, R$_2 = 8\Omega$, R$_3 = 12\Omega$, R$_4 = 24\Omega$.

(a) The highest total resistance is obtained when all resistors are connected in **series**. The equivalent resistance is the sum of individual resistances.

$R_{highest} = R_1 + R_2 + R_3 + R_4 = 4\Omega + 8\Omega + 12\Omega + 24\Omega = 48\Omega$.

The highest total resistance is 48 $\Omega$.

(b) The lowest total resistance is obtained when all resistors are connected in **parallel**. The reciprocal of the equivalent resistance is the sum of the reciprocals.

$\frac{1}{R_{lowest}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4\Omega} + \frac{1}{8\Omega} + \frac{1}{12\Omega} + \frac{1}{24\Omega}$.

Find a common denominator (24):

$\frac{1}{R_{lowest}} = \frac{6}{24} + \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6+3+2+1}{24} = \frac{12}{24} = \frac{1}{2}$.

$R_{lowest} = 2 \Omega$.

The lowest total resistance is 2 $\Omega$.

Given: Voltage, V = 220 V.

Power (rate of energy consumption) is given by P = VI or P = V$^2$/R.

From P = VI, current $I = P/V$. From P = V$^2$/R, resistance $R = V^2/P$ (or $R = V/I$).

**Case 1: Maximum heating rate**

Power, P$_{max}$ = 840 W.

Current, $I_{max} = P_{max} / V = 840\text{W} / 220\text{V} = 84/22\text{ A} = 42/11\text{ A} \approx 3.82\text{ A}$.

Resistance, $R_{max} = V / I_{max} = 220\text{V} / (42/11\text{ A}) = 220 \times 11 / 42 \Omega = 2420 / 42 \Omega = 1210 / 21 \Omega \approx 57.62 \Omega$.

Alternatively, $R_{max} = V^2 / P_{max} = (220\text{V})^2 / 840\text{W} = 48400 / 840 \Omega = 4840 / 84 \Omega = 1210 / 21 \Omega \approx 57.62 \Omega$.

Current is approximately 3.82 A, Resistance is approximately 57.62 $\Omega$ at maximum heating.

**Case 2: Minimum heating rate**

Power, P$_{min}$ = 360 W.

Current, $I_{min} = P_{min} / V = 360\text{W} / 220\text{V} = 36/22\text{ A} = 18/11\text{ A} \approx 1.64\text{ A}$.

Resistance, $R_{min} = V / I_{min} = 220\text{V} / (18/11\text{ A}) = 220 \times 11 / 18 \Omega = 2420 / 18 \Omega = 1210 / 9 \Omega \approx 134.44 \Omega$.

Alternatively, $R_{min} = V^2 / P_{min} = (220\text{V})^2 / 360\text{W} = 48400 / 360 \Omega = 4840 / 36 \Omega = 1210 / 9 \Omega \approx 134.44 \Omega$.

Current is approximately 1.64 A, Resistance is approximately 134.44 $\Omega$ at minimum heating. (Note: The iron changes its resistance to regulate heating power).

Given: Heat produced, H = 100 J.

Time, t = 1 second.

Resistance, R = 4 $\Omega$.

Find potential difference (V) across the resistor.

Using Joule's Law of Heating, $H = I^2Rt$, we can first find the current (I):

$100\text{ J} = I^2 \times (4\Omega) \times (1\text{ s})$.

$100 = 4 I^2$

$I^2 = 100 / 4 = 25$.

$I = \sqrt{25} = 5\text{ A}$ (current magnitude is positive).

Now, using Ohm's Law, $V = IR$, we can find the potential difference:

$V = (5\text{ A}) \times (4\Omega) = 20\text{ V}$.

The potential difference across the resistor is 20 V.

Both the cord and the heating element of an electric heater carry the same amount of current (since they are in series). However, the heating element is made of an alloy (like Nichrome) which has a much higher electrical resistance compared to the copper wires used in the cord. According to Joule's Law of Heating ($H = I^2Rt$), for the same current and time, the heat produced is directly proportional to the resistance. The heating element's high resistance causes it to generate a large amount of heat, making it get very hot and glow. The cord's low resistance generates much less heat, so it does not glow.

Given: Charge, Q = 96000 C.

Time, t = 1 hour. Convert to seconds: $t = 1 \times 3600 \text{ s} = 3600 \text{ s}$.

Potential difference, V = 50 V.

The heat generated (H) is equal to the electrical energy consumed, which is given by H = V $\times$ Q:

$H = (50\text{ V}) \times (96000\text{ C}) = 4800000\text{ J}$.

The heat generated is 4,800,000 J or $4.8 \times 10^6$ J.

Alternatively, we can find the current first: $I = Q/t = 96000\text{C} / 3600\text{s} = 960/36\text{ A} = 80/3\text{ A}$. Then use $H=VIt$: $H = (50\text{V}) \times (80/3\text{A}) \times (3600\text{s}) = 50 \times 80 \times 1200 \text{ J} = 4000 \times 1200 \text{ J} = 4800000\text{ J}$.

Given:

Resistance, R = 20 $\Omega$.

Current, I = 5 A.

Time, t = 30 s.

Using Joule's Law of Heating, $H = I^2Rt$:

$H = (5\text{ A})^2 \times (20\Omega) \times (30\text{ s})$

$H = 25 \times 20 \times 30 \text{ J}$

$H = 500 \times 30 \text{ J} = 15000 \text{ J}$.

The heat developed in 30 seconds is 15,000 J or 15 kJ.

Given: Voltage, V = 220 V.

Current, I = 0.50 A.

Power, P = ?

Using the formula P = VI:

$P = (220\text{ V}) \times (0.50\text{ A})$

$P = 110\text{ W}$.

The power of the electric bulb is 110 watts.

Given:

Power rating of refrigerator, P = 400 W.

Operating time per day = 8 hours.

Number of days = 30 days.

Cost per kWh = $\textsf{₹}$ 3.00.

First, calculate the total energy consumed by the refrigerator in 30 days. Energy consumed = Power $\times$ Total Time.

Total operating time = (8 hours/day) $\times$ (30 days) = 240 hours.

Energy consumed in Wh = Power (in W) $\times$ Total Time (in h) = 400 W $\times$ 240 h = 96000 Wh.

Convert energy to kWh: 1 kWh = 1000 Wh. So, Energy in kWh = 96000 Wh / 1000 Wh/kWh = 96 kWh.

Now, calculate the cost of energy:

Cost = Energy consumed (in kWh) $\times$ Cost per kWh

Cost = 96 kWh $\times$ $\textsf{₹}$ 3.00/kWh = $\textsf{₹}$ 288.00.

The cost of energy to operate the refrigerator for 30 days is $\textsf{₹}$ 288.00.

The rate at which energy is delivered by a current is determined by **electric power (P)**. Electric power is given by the formulas P = VI, P = I$^2$R, or P = V$^2$/R. It indicates how quickly electrical energy is converted into other forms of energy (like heat, light, mechanical energy) in a circuit.

Given: Current taken by motor, I = 5 A.

Voltage of the line, V = 220 V.

Time of operation, t = 2 hours.

Power of the motor (P):

Using the formula P = VI:

$P = (220\text{ V}) \times (5\text{ A}) = 1100\text{ W}$.

The power of the motor is 1100 watts (or 1.1 kW).

Energy consumed (E) in 2 hours:

Using the formula Energy = Power $\times$ Time. We can calculate in Wh or kWh.

Energy in Wh = Power (in W) $\times$ Time (in h) = 1100 W $\times$ 2 h = 2200 Wh.

Convert to kWh: 2200 Wh / 1000 Wh/kWh = 2.2 kWh.

Convert to Joules: 1 kWh = $3.6 \times 10^6$ J. So, 2.2 kWh $\times$ $3.6 \times 10^6$ J/kWh = $7.92 \times 10^6$ J.

The energy consumed by the motor in 2 hours is 2.2 kWh or $7.92 \times 10^6$ J.