(a) Two lamps rated $100 \text{ W}$, $220 \text{ V}$ and $10 \text{ W}$, $220 \text{ V}$ are connected in parallel to $220 \text{ V}$ supply. Calculate the total current through the circuit.

(b) Two resistors X and Y of resistances $2 \Omega$ and $3 \Omega$ respectively are first joined in parallel and then in series. In each case the voltage supplied is $5 \text{ V}$.

(i) Draw circuit diagrams to show the combination of resistors in each case.

(ii) Calculate the voltage across the $3 \Omega$ resistor in the series combination of resistors.

(a) Calculation of Total Current through the Parallel Circuit

Given:

• Power of the first lamp, $P_1 = 100 \text{ W}$
• Power of the second lamp, $P_2 = 10 \text{ W}$
• Common potential difference (voltage), $V = 220 \text{ V}$

Since the two lamps are connected in parallel across the same voltage source of $220 \text{ V}$, the potential difference across each lamp is $220 \text{ V}$.

We know that electric power is given by:

$$P = V \times I$$

Therefore, the current drawn by each lamp is:

$$I_1 = \frac{P_1}{V} = \frac{100 \text{ W}}{220 \text{ V}} = \frac{5}{11} \text{ A} \approx 0.455 \text{ A}$$

$$I_2 = \frac{P_2}{V} = \frac{10 \text{ W}}{220 \text{ V}} = \frac{1}{11} \text{ A} \approx 0.045 \text{ A}$$

In a parallel circuit, the total current $I$ is the sum of individual currents flowing through each branch:

$$I = I_1 + I_2$$

$$I = \frac{5}{11} \text{ A} + \frac{1}{11} \text{ A} = \frac{6}{11} \text{ A} \approx 0.55 \text{ A}$$

(Alternatively, total power $P_{\text{total}} = P_1 + P_2 = 100 \text{ W} + 10 \text{ W} = 110 \text{ W}$. Total current $I = \frac{P_{\text{total}}}{V} = \frac{110 \text{ W}}{220 \text{ V}} = 0.5 \text{ A}$.)

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(b) Resistors X and Y in Parallel and Series

#### (i) Circuit Diagrams

1. Series Combination:

In a series combination, the resistors are connected end-to-end so that the same current flows through them.

2. Parallel Combination:

In a parallel combination, the resistors are connected across the same two common nodes, keeping the potential difference across them equal.

#### (ii) Calculation of Voltage across the $3 \Omega$ Resistor in Series

1. Equivalent Resistance in Series ($R_s$):

$$R_s = R_X + R_Y = 2 \; \Omega + 3 \; \Omega = 5 \; \Omega$$

1. Total Current in the Circuit ($I$):

Using Ohm's law ($V = I R_s$):

$$I = \frac{V}{R_s} = \frac{5 \text{ V}}{5 \; \Omega} = 1 \text{ A}$$

1. Voltage across the $3 \; \Omega$ Resistor ($V_Y$):

Since the same current of $1 \text{ A}$ flows through both resistors in series:

$$V_Y = I \times R_Y = 1 \text{ A} \times 3 \; \Omega = 3 \text{ V}$$

Thus, the voltage across the $3 \; \Omega$ resistor is $3 \text{ V}$.