Question Details

Choose the correct circuit which can achieve the bridge balance

Options

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Correct Answer :

Option A

Option 1

Solution :

The correct circuit is shown in Option 1 (the first diagram).

1. Condition for Bridge Balance:
A Wheatstone bridge is balanced when the potential at the top junction (connected to the galvanometer) is equal to the potential at the bottom junction, resulting in zero current flowing through the galvanometer $G$ :

$V_{\text{top}} = V_{\text{bottom}}$

Let the left junction (connected to the positive terminal of the cell $E$ ) be at potential $V_{\text{A}}$ , and the right junction (connected to the negative terminal through key $K$ ) be at potential $V_{\text{B}}$ , where $V_{\text{A}} > V_{\text{B}}$ .

2. Potential at the Top Junction:
The upper branch consists of two series resistors of $10 \Omega$ each. Using the voltage divider rule, the potential at the top junction is:

$V_{\text{top}} = V_{\text{B}} + (V_{\text{A}} - V_{\text{B}}) \cdot \frac{10}{10 + 10} = \frac{V_{\text{A}} + V_{\text{B}}}{2}$

3. Potential at the Bottom Junction:
For the bridge to be balanced, the potential at the bottom junction must also satisfy:

$V_{\text{bottom}} = \frac{V_{\text{A}} + V_{\text{B}}}{2}$

This requires the resistance of the bottom-left arm ( $15 \Omega$ ) to be equal to the total effective resistance of the bottom-right arm ( $R_{\text{bottom-right}}$ ):

$R_{\text{bottom-right}} = 15 \Omega$

4. Analyzing the Bottom-Right Arm with the Diode:
The bottom-right arm contains a $5 \Omega$ resistor and a diode $D$ connected in series. For the branch to have a total resistance of $15 \Omega$ , the diode must conduct and present a finite dynamic resistance ( $R_{\text{diode}}$ ):

$5 \Omega + R_{\text{diode}} = 15 \Omega \Rightarrow R_{\text{diode}} = 10 \Omega$

For the diode to conduct and exhibit this dynamic resistance, it must be forward-biased. Since current flows from the higher potential at the bottom junction to the lower potential at the right junction, the diode's anode must be connected towards the bottom junction and its cathode towards the right junction. This means the diode symbol's arrow (triangle) must point towards the right node (top-right), which is the configuration shown in the first circuit (Option 1).