Question

Two resistances each of 10 ohm are connected in parallel. A battery of 10 V is connected with this combination. Amount of charge flowing per second will be -

• A. 4 coulomb/second
• B. 2 coulomb/second
• C. 1 coulomb/second
• D. 3 coulomb/second

Hint:

In a parallel combination of resistors, the equivalent resistance is less than the smallest individual resistance. The total current can then be calculated using Ohm's Law.

Answer 1

The Correct Option is B

Step 1: Understanding the situation
Two resistances \( R_1 \) and \( R_2 \) of 10 ohm each are connected in parallel, and a battery of 10 V is connected across them. The total current flowing through the combination is calculated using Ohm's Law and the formula for parallel resistances.

Step 2: Find the equivalent resistance
For two resistances in parallel, the equivalent resistance \( R_{eq} \) is given by: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \] Substitute \( R_1 = R_2 = 10 \, \Omega \): \[ \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \] So, the equivalent resistance is: \[ R_{eq} = 5 \, \Omega \]

Step 3: Apply Ohm's Law to find the current
Ohm's law states: \[ I = \frac{V}{R_{eq}} \] Substitute \( V = 10 \, \text{V} \) and \( R_{eq} = 5 \, \Omega \): \[ I = \frac{10}{5} = 2 \, \text{A} \] Thus, the current flowing through the circuit is \( 2 \, \text{A} \).

Step 4: Calculate the amount of charge flowing per second
The current is defined as the rate of flow of charge: \[ I = \frac{Q}{t} \] Where \( Q \) is the charge and \( t \) is the time. Since the current is \( 2 \, \text{A} \), it means that 2 coulombs of charge flow per second. Thus, the charge flowing per second is \( 2 \, \text{coulombs} \).
The correct answer is option (B) \( 2 \, \text{coulombs/second} \).