1: Analyzing the Circuit
The given circuit consists of resistors in series and parallel. First, simplify the resistances step by step. 1. Combine the 10\(\mu\) and 10\(\mu\) resistors in parallel between points B and C: \[ R_{\text{BC}} = \frac{1}{\frac{1}{10} + \frac{1}{10}} = 5 \] 2. Combine the 20\(\mu\) resistors in parallel between points C and D: \[ R_{\text{CD}} = \frac{1}{\frac{1}{20} + \frac{1}{20}} = 10 \] 3. Simplify the overall network of resistors to find the effective resistance between points A and M: \[ R_{\text{eff}} = 12 \]
2: Power Supplied by the Battery
The power supplied by the battery is given by: \[ P = \frac{V^2}{R_{\text{eff}}} = \frac{6^2}{12} = 3W \] Thus, the power supplied by the battery is \( P = 3W \).
The current passing through the battery in the given circuit, is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :