A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in

series across a 90.0 V battery. What is the current in the

circuit?

a 5A

b

10 A.

с ЗА

d 8A

Answer: Option C) 3A

Explanation:

Given that,

Voltage = 90.0 Volts

Current I = ?

R1 = 5.0Ω

R2 = 10.0Ω

R3 = 15.0Ω

Since the resistors are connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i. e Rtotal = R1 + R2 + R3

Rtotal = 5.0Ω + 10.0Ω + 15.0Ω = 30.0Ω

Then, apply the formula for ohms law

Voltage = Current x resistance

V = I x Rtotal

90.0V = I x 30.0Ω

I = 90.0V / 30.0Ω

I = 3 Amps (Amps is the unit of current)

Thus, the current in the circuit is 3 Amps