
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.
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at \(V_z = 5\) V and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25 V. Considering the Zener diode can withstand four times of the load current, the value of resistor \(R_s\) (shown in circuit) should be_______ \(\Omega\).
An object is projected with kinetic energy K from point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies at points B and C to that at point A (see figure), in the absence of air friction is : 