Applying Kirchhoff’s Laws
Given that the equivalent resistance across CH is: \[ R_{CH} = 2R \] Equivalent Circuit Diagram
The given circuit is analyzed using Kirchhoff’s Voltage Law (KVL). We apply KVL to two loops in the circuit. Applying Kirchhoff’s Voltage Law (KVL):
In closed loop ABMNA:
Using KVL in loop ABMNA, we sum the voltage drops: \[ -3IR - 4I_1 R + 16E = 0 \]
This forms our first equation: -3IR - 4I_1 R + 16E = 0
In closed loop BCHMB:
Similarly, applying KVL in loop BCHMB: \[ -2R(I - I_1) - 6E + 4I_1 R = 0 \] This forms our second equation: -2R(I - I_1) - 6E + 4I_1 R = 0
Solving the Equations
Now, solving equations (1) and (2) simultaneously, we obtain: \[ I_1 = \frac{25E}{13R} \] Thus, the current \( I_1 \) in the circuit is determined.
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 :