Four resistors are connected to a battery as shown in the fi

Four resistors are connected to a battery as shown in the figure. The current in the battery is t, the battery is epsilon = 11.10 V, and the resistor value are R_1 = R_2

Solution

While solving any circuit, we need to keep following points into consideration:

1.) The current in the resistors in parallel remains the same.

2.) The voltage across two lines parallel to each other is same.

3.) For resistors R1 and R2 in series, the net resistance is given as R1 + R2, while for parallel connection, the net resistance is 1 / (1/R1 + 1/R2)

Now, for the given circuit,net resistance for R2 and R3 is 6R

while for R2, R3 and R4 would be 6 x 3 / 9 = 2R, hence the net resistance of the circuit would be 3R

that is I = V / 3R

So the voltage across R1 = RI =R x V / 3R = V/3 = 11.1 / 3 = 3.7 Volts

Voltage across R2, R3 and R4 would be: I x Rnet = 2R *V / 3R = 2V/3

So the voltage across, R2 and R3 is 2V/3, hence, I(Through R2 and R3) = 2V/3*6R = V/9R

That is voltage across R2 = 2RV/9R = 2V/9 = 2.467 Volts

Voltage acorss R3 = 4R V/9R = 4V/9 = 4.93 Volts

Voltage across R4 = 2V/3 = 7.4 Volts