A battery is supplying power to a tape recorder by cable of resistance of `0.02 Omega`. If the battery is generating 50 W power at 5 V, then the power received by the tape recorder is

Options

A. 50 W
B. 45 W
C. 30 W
D. 48 W

Explanation

To solve the problem, we need to find the power received by the tape recorder when a battery is supplying power through a cable with a given resistance. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Power generated by the battery (P_battery) = 50 W
- Voltage of the battery (V) = 5 V
- Resistance of the cable (R_cable) = 0.02 Ω

### Step 2: Calculate the current flowing through the circuit
Using the formula for power:
\[ P = V \times I \]
We can rearrange this to find the current (I):
\[ I = \frac{P}{V} \]
Substituting the values:
\[ I = \frac{50 \, \text{W}}{5 \, \text{V}} = 10 \, \text{A} \]

### Step 3: Calculate the power lost in the cable
The power lost in the cable due to its resistance can be calculated using the formula:
\[ P_{loss} = I^2 \times R \]
Substituting the values:
\[ P_{loss} = (10 \, \text{A})^2 \times 0.02 \, \Omega = 100 \times 0.02 = 2 \, \text{W} \]

### Step 4: Calculate the power received by the tape recorder
The power received by the tape recorder can be found by subtracting the power lost in the cable from the total power generated by the battery:
\[ P_{tape\ recorder} = P_{battery} - P_{loss} \]
Substituting the values:
\[ P_{tape\ recorder} = 50 \, \text{W} - 2 \, \text{W} = 48 \, \text{W} \]

### Final Answer
The power received by the tape recorder is **48 W**.

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