Question

Three resistors $R_1, R_2$ and $R_3$ when connected in parallel to a battery of negligible internal resistance, the currents through the three resistors are 2A, 3A and 6A respectively. If the resistors are connected in series to the same battery, the current in the circuit is

• A. 1A
• B. 2A
• C. 3A
• D. 11A

Hint:

Using proportionality helps. Resistances are proportional to inverse of currents in parallel. $R \propto 1/I$.

Answer 1

The Correct Option is A

Step 1: Parallel Connection Analysis:
Let the battery voltage be $V$. In parallel, voltage across each resistor is $V$. $R_1 = V/I_1 = V/2$. $R_2 = V/I_2 = V/3$. $R_3 = V/I_3 = V/6$.
Step 2: Series Connection Analysis:
Equivalent resistance in series: $R_{eq} = R_1 + R_2 + R_3$. \[ R_{eq} = \frac{V}{2} + \frac{V}{3} + \frac{V}{6} \] \[ R_{eq} = V \left( \frac{3 + 2 + 1}{6} \right) = V \left( \frac{6}{6} \right) = V \]
Step 3: Calculate Series Current:
Current $I_s = \frac{V}{R_{eq}}$. \[ I_s = \frac{V}{V} = 1 \, A \]
Step 4: Final Answer:
The current is 1 A.