The current source with internal resistance and EMF is closed by three resistors with a resistance of 3r each connected in series.

The current source with internal resistance and EMF is closed by three resistors with a resistance of 3r each connected in series. How many times will the current in the circuit and the voltage at the source terminals change if the resistors are connected in parallel?

Find the load resistance in series connection
Rn1 = 3r + 3r + 3r = 9r.

Total circuit resistance
Ro1 = r + Rn1 = r + 9r = 10r.

Let’s designate the EMF of the source through E and write the expression for the current
I1 = E / Ro1 = E / 10r.

Load voltage
U1 = I1 * Rn1 = (E / 10r) * 9r = 9E / 10.

For parallel connection
Rn2 = 3r / 3 = r;

Ro2 = r + Rn2 = r + r = 2r;
I2 = E / Ro2 = E / 2r;
U2 = I2 * Rn2 = E * r / 2r = E / 2.

The ratio of currents and voltages
I2 / I1 = (E / 2r) / (E / 10r) = 10/2 = 5;
U1 / U2 = (9E / 10) / (E / 2) = 2 * 9/10 = 18/10 = 1.8.

Answer: the current will change 5 times, the voltage 1.8 times.