Electrical Circuit Basics

Question

When we connect an R resistor to a battery battery, the resistor's ends voltage is measured at 240 V. When we add 10 Ω to this resistance, the circuit current becomes 5 A, and when 20 Ω is removed, it becomes 10 A. Calculate the voltage of the battery, the internal resistance of the battery, R and I.

Answer 1

From R_internal = V_battery/I − R, write:

(1) R_int = V_battery/I − (R + 0);

(2) R_int = V_battery/(5 A) − (R + 10);

(3) R_int = V_battery/(10 A) − (R − 20).

Multiply I through Equation (1) to obtain IR_int = V_battery − IR or IR_int = V_battery − 240.

With V_battery = [IR_int + 240], use Equations (2) & (3) to arrive at

[IR_int + 240]/(5 A) − R − 10 = [IR_int + 240]/(10 A) − R + 20.

Next write 10{[IR_int + 240]/(5 A) − R − 10 = [IR_int + 240]/(10 A) − R + 20} which gives

2IR_int + 480 − 10R − 100 = IR_int + 240 − 10R + 200. Simplify further to

IR_int + 380 = 440 or IR_int = 60 Volts. From V_battery equal to IR_int + V_R, write

V_battery = 60 Volts + 240 Volts or 300 Volts.

With guidance from Equations (1), (2), & (3), the information above will give:

60 = 300 − 240;

60 = 60I − 240 − 10I;

60 = 30I − 240 + 20I.

The first equation being trivial, the second and third equations both yield 60 = 50I − 240 or

I = 6 Ampères.

Next, IR = 6R equal to 240 Volts gives R = 40 Ohms.

Finally, IR_int = 6R_int = 60 Volts, so R_int = 10 Ohms.