Three resistors Of 5 Ω, 10 Ω and 15 Ω are connected in series with a 12 V power supply.

(a) Calculate their combined resistance and the current that flows in the circuit and in each resistor.

(b) Find the potential difference across each resistor.

(a) In the circuit 5 Ω, 10 Ω and 15 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 5 + 10 + 15 = 30 Ω

According to Ohm's law: V = IR

Substituting the values in the formula:

12 = I x 30

I = $\dfrac{12}{30}$ = 0.4 A

Hence, combined resistance = 30 Ω and current flowing in the circuit = 0.4 A

As in a series circuit current flowing through each resistor is equal to the total current flowing in the circuit, hence, current flowing through each resistor = 0.4 A

(b) Potential difference across 5 Ω resistor: ?

Applying Ohm's law: V = IR

Substituting the values in the formula:

V = 0.4 x 5 = 2 V

Potential difference across 10 Ω resistor: ?

V = 0.4 x 10 = 4 V

Potential difference across 15 Ω : ?

V = 0.4 x 15 = 6 V

Hence, potential difference across 5 Ω resistor = 2, 10 Ω resistor = 4V, 15 Ω resistor = 6V