Potential difference between the points A and B is nearly
- 10 V
- 14 V
- 18 V
- 20 V
Solution
The given circuit is a Wheatstone's bridge, As the resistances are in the same ratio, the bridge is balanced. Therefore, no current will flow through the galvanometer and the resistances will be in parallel combination. $\begin{array}{ll} \therefore \quad & \text { Voltage between } A \text { and } C \text { is } V_{A C}=I . R_{\text {eff }} \\ & \frac{1}{R_{\text {eff }}}=\frac{1}{(8+4)}+\frac{1}{(10+5)} \Rightarrow R_{\text {eff. }}=\frac{15 \times 12}{27}=6.67 \Omega \\ \therefore \quad & V_{A C}=4 \times 6.67=26.68 \mathrm{~V} ...[From(i)]\\ \therefore & \mathrm{I}_{A C}=\frac{V_{A C}}{R_{A B}+R_{B C}}=\frac{26.68}{8+4}=2.22 \mathrm{~A} \\ \therefore & V_{A B}=I_{A C} \times R_{A B}=2.22 \times 8=17.76 \mathrm{~V}=18 \mathrm{~V} \end{array}$
Asked in: MHT CET 2024 (09 May Shift 2)