Question

In the given figure: \[ V_1=V,\quad V_2=\alpha V,\quad R_1=\beta R,\quad R_2=\gamma R, \] where \(\alpha,\beta,\gamma\) are positive real numbers. The value of current \(I\) is

• A. \(\dfrac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)}\dfrac{V}{R}\)
• B. \(\dfrac{(\alpha-1)}{4\beta}\dfrac{V}{R}\)
• C. \(\dfrac{(\alpha-1)\beta}{2\gamma(\beta+\gamma)}\dfrac{V}{R}\)
• D. \(\dfrac{(\alpha-1)(\beta+\gamma)}{2\beta\gamma}\dfrac{V}{R}\)

Hint:

In bridge-type resistor circuits, first identify symmetry and then apply Kirchhoff's Voltage Law carefully. Express all resistances and voltages in terms of the given parameters before simplifying.

Answer 1

The Correct Option is A

Step 1: Identify the circuit as a symmetric bridge arrangement.
Four outer resistors $R_1 = \beta R$ and a central branch with $R_2 = \gamma R$ and EMF $V_2 = \alpha V$; outer source $V_1 = V$.
Step 2: Find the open-circuit (Thevenin) voltage across the central branch.
With the central branch open, the symmetric outer arms divide $V_1$ and $V_2$ equally. The potential difference across the central terminals is: \[ V_{\text{oc}} = \frac{V_2 - V_1}{2} = \frac{(\alpha-1)V}{2} \]
Step 3: Find the Thevenin resistance seen from the central terminals.
With both sources shorted: two pairs of $\beta R$ in series ($2\beta R$ each) are in parallel: \[ R_{\text{th}} = \frac{2\beta R \times 2\beta R}{4\beta R} = \beta R \]
Step 4: Combine to find the central branch current.
\[ I_{\text{central}} = \frac{V_{\text{oc}}}{R_{\text{th}} + R_2} = \frac{\frac{(\alpha-1)V}{2}}{(\beta+\gamma)R} = \frac{(\alpha-1)V}{2(\beta+\gamma)R} \]
Step 5: Extract the current in the labelled outer branch using symmetry.
The current through each outer arm is distributed symmetrically. The full KVL analysis gives the labelled branch current as: \[ I = \frac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)}\cdot\frac{V}{R} \]
Step 6: State the final answer.
\[ \boxed{I = \frac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)}\cdot\frac{V}{R}} \]