In the given circuit, batteries are ideal and Galvanometer shows zero deflection, then the value of 'R' is:
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For a balanced Wheatstone bridge, ensure all resistances are correctly identified, and the balance condition formula is properly applied. Misinterpretations or incorrect assumptions can lead to errors in calculations.
For the galvanometer to show zero deflection, the potential difference across its terminals must be zero. This implies that the potential at point X must be equal to the potential at point Y. Let's assume the negative terminals of both batteries are at ground potential (0V). The potential at the positive terminal of the 2V battery is 2V relative to its negative terminal. Thus, the potential at point Y is 2V. For zero deflection, the potential at point X must also be 2V. Point X is in a voltage divider circuit formed by the 300Ω resistor and resistor R, connected to the 12V battery. Using the voltage divider rule, the voltage across resistor R (which is the potential at point X relative to the negative terminal of the 12V battery) is given by: \[ V_R = V_{12V} \times \frac{R}{R + 300} \] Where \(V_{12V} = 12V\) and we require \(V_R = 2V\) for zero deflection. Substituting these values: \[ 2 = 12 \times \frac{R}{R + 300} \] To solve for R, we first divide both sides by 2: \[ 1 = 6 \times \frac{R}{R + 300} \] Multiply both sides by \(R + 300\): \[ R + 300 = 6R \] Rearrange the equation to solve for R: \[ 300 = 6R - R \] \[ 300 = 5R \] \[ R = \frac{300}{5} \] \[ R = 60 \, \Omega \] Therefore, the value of R for zero galvanometer deflection is \(60 \, \Omega\). Correct Answer: (1) \(60 \, \Omega\)
