In the circuit shown in the figure below the batteries have

In the circuit shown in the figure below, the batteries have negligible internal resistance. Find the current in each branch of the circuit. I_1ohm = A I_2ohm = A I_3ohm = A Find the potential difference between point a and point b. V Find the power supplied by each battery. P_5V = W P_7V = W

Solution

Let \"I₂\" be current delivered by the 7 Volt battery ,\"I₃\" be current delivered by the 5 Volt battery and \"I₁\" be current through 1 ohms resistor.

From Kirchoff\'s Junction rule

I₂ = I₃+I₁

I₁-I₂+I₃=0------------------1

Applying kirchoff\'s loop rule to outer loop of the circuit

7-2I₂-I₁=0

I₁+2I₂=7----------------------2

Applying kirchoff\'s loop rule to left loop of the circuit

7-2I₂-3I₃+5=0

2I₂+3I₃=12----------------------------3

Solving 1,2 and 3 we get

I₁=1 A

I₂=3 A

I₃=2 A

b)

Potential difference between terminals \"a\" and \"b\" is

V_ab=-5+3I₃ =-5+3*2 =1 Volts

c)

Power delivered by 7 Volt battery is

P₇=3*7 =21 Watts

Power delivered by 5 Volt battery is

P₅=2*5 =10 Watts