In a conductor with a resistance of 8 Ohm connected to a source with an EMF18V, a power of 32 W

In a conductor with a resistance of 8 Ohm connected to a source with an EMF18V, a power of 32 W is released. What is the strength of the short circuit current of the source.

The allocated power P through the current I and the resistance of the conductor R is determined by the formula:

P = I2R;

I = √ (P / R) = √32 / 8 = 2 A.

Current in circuit I:

I = E / (R + r),

where E is the EMF of the source, r is the internal resistance.

IR + Ir = E;

r = (E – IR) / I = E / I – R = 18 V / 2 A – 8 Ohm = 9 Ohm – 8 Ohm = 1 Ohm.

Short-circuit current Isc = E / r = 18 V / 1 Ohm = 18 A.

Answer: 18 A.