#### Question

For the circuit shown in the diagram below:

What is the value of:

(i) current through 6 Ω resistor?

(ii) potential difference across 12 Ω resistor?

#### Solution

The resistors of 6 Ω and 3 Ω are connected in series. Therefore, their net resistance can be calculated as:

*R* = *R*_{1} + *R*_{2}

Here, *R*_{1}_{ }= 6 Ω

*R*_{2} = 3 Ω

So:*R* = 6 Ω + 3 Ω = 9 Ω

The current through this branch, *I* = *V*/*R**I* = 4/9 = 0.44 A

In a series combination, the current remains the same. So the current through the 6 Ω resistor is 0.44 A.

(2) The current through the branch with resistors of 12 Ω and 3 Ω:*I* = *V*/*R**I* = 4/(12 + 3) = 4 / 15 A

The potential difference across the 12 Ω resistance can be obtained by using the equation,*V* = *IR*.*V* = (4 / 15) x 12 = 3.2 V

Is there an error in this question or solution?

Solution For the Circuit Shown in the Diagram Below: Concept: Electric Potential Difference.