Question

Two resistors of values 7.0 and 15.0 Ω are connected in parallel. This combination in turn is hooked in series with a 3.8- Ω resistor and a 22.3-V battery. What is the current in the 3.8- Ω resistor?

Answer 1

1.8A

Step-by-step explanation:

Firstly, we need to find the effective resistance in the circuit.

For the two parallel resistors, the effective resistance is;

1/R = 1/7 + 1/15

R = 8.57Ω

Now let's connect it in series with the series resistor

R = 8.57 + 3.8 = 12.37Ω

This is the total effective resistance

The total current is therefore

22.3/8.57 = 1.8A

This is also the current flowing through the 3.8 resistor since it is connected in series