Three electric bulbs rated 40, 60 and 100 watts respectively are connected in parallel with 230 volts d.c. mains. The current (in amp) through the 60 W bulb and the effective resistance (in ohm) of the circuit will be respectively, approximately

Options

A. `0.23 A`
B. `329.Omega`
C. `0.26 A`
D. `265Omega`

Answer

To solve the problem step by step, we will first calculate the current flowing through the 60 W bulb and then find the effective resistance of the circuit.

### Step 1: Calculate the Resistance of the 60 W Bulb
The power \( P \) of the bulb can be expressed in terms of voltage \( V \) and resistance \( R \) using the formula:
\[ P = \frac{V^2}{R} \]
Rearranging this gives us:
\[ R = \frac{V^2}{P} \]
For the 60 W bulb connected to a 230 V supply:
\[ R_2 = \frac{230^2}{60} \]
Calculating this:
\[ R_2 = \frac{52900}{60} \approx 881.67 \, \Omega \]

### Step 2: Calculate the Current Through the 60 W Bulb
Using Ohm's law, the current \( I \) through the bulb can be calculated as:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I_2 = \frac{230}{R_2} = \frac{230}{881.67} \approx 0.26 \, \text{A} \]

### Step 3: Calculate the Effective Resistance of the Circuit
Since the bulbs are connected in parallel, the effective resistance \( R_{eq} \) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Where:
- \( R_1 = \frac{230^2}{40} \)
- \( R_2 = \frac{230^2}{60} \)
- \( R_3 = \frac{230^2}{100} \)

Calculating each resistance:
\[ R_1 = \frac{52900}{40} \approx 1322.5 \, \Omega \]
\[ R_2 = \frac{52900}{60} \approx 881.67 \, \Omega \]
\[ R_3 = \frac{52900}{100} \approx 529 \, \Omega \]

Now substituting these into the effective resistance formula:
\[ \frac{1}{R_{eq}} = \frac{1}{1322.5} + \frac{1}{881.67} + \frac{1}{529} \]
Calculating each term:
\[ \frac{1}{R_{eq}} \approx 0.000755 + 0.001134 + 0.001891 \approx 0.00378 \]
Thus,
\[ R_{eq} \approx \frac{1}{0.00378} \approx 264.55 \, \Omega \approx 265 \, \Omega \]

### Final Answers
- The current through the 60 W bulb is approximately **0.26 A**.
- The effective resistance of the circuit is approximately **265 Ω**.

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