Four resistence of `10 Omega, 60Omega,100Omega`and `200Omega`, respectively taken in order are used to form a Wheatstone's bridge . A 15 V battery is connected to the ends of a `200Omega`resistance, the current through it will be

Options

A. `7.5xx10^(-5)A`
B. `7.5xx10^(-4)A`
C. `7.5xx10^(-3)A`
D. `7.5xx10^(-2)A`

Answer

To find the current through the 200 Ω resistor in a Wheatstone bridge with the given resistances, we can follow these steps:

### Step 1: Identify the total voltage across the 200 Ω resistor
We know that a 15 V battery is connected across the 200 Ω resistor. Therefore, the potential difference (V) across the 200 Ω resistor is 15 V.

**Hint:** Remember that the voltage provided by the battery is the potential difference across the resistor in question.

### Step 2: Use Ohm's Law to calculate the current
Ohm's Law states that \( V = I \cdot R \), where:
- \( V \) is the voltage across the resistor,
- \( I \) is the current through the resistor,
- \( R \) is the resistance.

We can rearrange this formula to find the current:
\[ I = \frac{V}{R} \]

In this case, \( V = 15 \, \text{V} \) and \( R = 200 \, \Omega \).

### Step 3: Substitute the values into the equation
Now, substituting the values into the equation:
\[ I = \frac{15 \, \text{V}}{200 \, \Omega} \]

### Step 4: Calculate the current
Calculating the above expression:
\[ I = \frac{15}{200} = 0.075 \, \text{A} \]

### Step 5: Convert to milliamperes (if needed)
To express the current in milliamperes (mA):
\[ I = 0.075 \, \text{A} = 75 \, \text{mA} \]
Or in scientific notation:
\[ I = 7.5 \times 10^{-2} \, \text{A} \]

### Conclusion
The current through the 200 Ω resistor is \( 7.5 \times 10^{-2} \, \text{A} \) or \( 75 \, \text{mA} \).

**Final Answer:** \( 7.5 \times 10^{-2} \, \text{A} \) (Option D is correct).

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