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.
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process. \text{In the light of the above statements, choose the correct answer from the options given below:}
For the circuit shown above, the equivalent gate is:
Find the equivalent resistance between two ends of the following circuit:
The circuit consists of three resistors, two of \(\frac{r}{3}\) in series connected in parallel with another resistor of \(r\).