Calculate the amount of charge flowing in 2 minutes in a wire of resistance `10 Omega` when a potential difference of 20 V is applied between its ends

Options

A. 120 C
B. 240 C
C. 20 C
D. 4 C

Answer

To solve the problem of calculating the amount of charge flowing in 2 minutes in a wire of resistance \(10 \, \Omega\) when a potential difference of \(20 \, V\) is applied, we can follow these steps:

### Step 1: Use Ohm's Law to find the current (I)
Ohm's Law states that:
\[ V = I \cdot R \]
Where:
- \(V\) is the potential difference (voltage),
- \(I\) is the current,
- \(R\) is the resistance.

Given:
- \(V = 20 \, V\)
- \(R = 10 \, \Omega\)

We can rearrange the formula to solve for current \(I\):
\[ I = \frac{V}{R} \]

Substituting the given values:
\[ I = \frac{20 \, V}{10 \, \Omega} = 2 \, A \]

### Step 2: Convert time from minutes to seconds
The time given is \(2\) minutes. We need to convert this into seconds because the standard unit of time in physics is seconds.
\[ T = 2 \, \text{minutes} \times 60 \, \text{seconds/minute} = 120 \, \text{seconds} \]

### Step 3: Calculate the charge (Q)
The relationship between charge \(Q\), current \(I\), and time \(T\) is given by:
\[ Q = I \cdot T \]

Substituting the values we found:
\[ Q = 2 \, A \cdot 120 \, s = 240 \, C \]

### Final Answer
The amount of charge flowing in 2 minutes is:
\[ Q = 240 \, C \]

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