Three resistors and two batteries are connected as shown in the circuit diagram.

Question

Three resistors and two batteries are connected as shown in the circuit diagram. What it the magnitude of the current through the 12-V battery? Mark on the drawing the currents i1, i2, i3 and show their direction. When you calculate the 3 currents, use the junction rule to provide that they are correct.

Explanation / Answer

Let the current in the upper loop be i1 and that in the lower loop be i2 both in clockwise direction so 16i1 - 18 + 25 i1 + 12 = 0 => i1 = .146 A for the lower loop 18 i2 = 12 => i2 = .667 A Current through the 12 V battery = .667 - .146 = .521 A Current through the 18 V battery = .146 A in the right Current through the 12 V battery = .521 A in the right Current through the 18 Ohms resistor = .667 A to the left by junction rule we get .667 = .521 + .146 hence proved

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