Two resistors of 5.0 and 9.0 ohm are connected in parallel. A 4.0 ohm resistor is then connected in series with the parallel combination. A 6.0-V battery is then connected to the series-parallel combination. What is the current through the 5.0 ohm resistor?

I1 = 0.772 A

Explanation:

Given: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts

To find: current I = ? A

Solution:

Ohm's law V = I R

⇒ I = V / R

In order to find R (total) we first find R (p) fro parallel combination. so

1 / R (p) = 1 / R1 + 1 / R2 ∴ (P) stand for parallel

R (p) = R1R2 / (R1 + R2)

R (p) = (5.0 * 9.0) / (5.0 + 9.0)

R (p) = 3.214 ohm

Now R (total) = R (p) + R3 (as R3 is connected in series)

R (total) = 3.214 ohm + 4.0 Ohm

R (total) = 7.214 ohm

now I (total) = 7.214 ohm / 6.0 Volts

I (total) = 1.202 A

This the total current supplied by 6 volts battery.

as voltage drop across R (p) = V = R (p) * I (total)

V (p) = 3.214 ohm * 1.202 A = 3.864 volts

Now current through 5 ohms resister is I1 = V (P) / R1

I1 = 3.864 volts / 5 ohm

I1 = 0.772 A