Question

Part A: For the circuit shown in the figure(Figure 1) find the current through each resistor.

Part B: For the circuit shown in the figure find the potential difference across each resistor.

Figure 1 of 1 凰 24 V 8Ω 6Ω 12Ω

Answer 1

First of all we will calculate the equivalent resistance of the circuit
So 6 and 12 ohm are in parallel
There equivalent resistance will be
R = 12*6 / 12+6 = 4 ohm
This equivalent resistance will be in series with the 4 ohm
Hence the net resistance will be = 4+ 4 = 8 ohm.
This 8 ohm will be in parallel with the 8 ohm in last branch.
hence the net equivalent resistance will be = 8*8 / (8+8) = 4 ohm
And this will be in series with the 2 ohm.
hence the net equivalent resistance of the circuit = 2+ 4 = 6 ohm.

Now the current in the circuit
I = V/R = 24 / 6 = 4 A
Now current through 2 ohm resistor = 4 A and the voltage across it = IR = 4*2 = 8 V
Now voltage at point A after 2 ohm will be the 24 - 8 = 16 V
Now the voltage drop in the last two branch will be 16 V as they both are in parallel.
Hence I₁*8 = 16
I₁ = 2 A
Current through 8 ohm resistor = 2 A and the voltage = 2*8 = 16 V
Now At node B
I = I₁+ I₂
4 = 2 + I₂
I₂ = 2 A
Hence the current through 4 ohm resistor = 2 A and the voltage = 2*4 = 8 V
Since the voltage drop this branch was 16 V and the voltage lost in 4 ohm is 8 V therefore the voltage lost in 6 and 12 ohm will be = 16 - 8 = 8 V
Now the current in 6 ohm , I₃ = 8 / 6 = 1.333 A and the voltage is 8 V
Now the current in 12 ohm resistor = 8 /12 = 0.67 A and the voltgae is 8 V.