The circuit shown in the figure is connected to an EMF source of 4.5 V and an internal resistance of 1.5 Ohm.

univerkov | February 24, 2021 | education

The circuit shown in the figure is connected to an EMF source of 4.5 V and an internal resistance of 1.5 Ohm. What is the current strength in the unbranched part of the circuit if R1 = R2 = 10 ohms, R3 = 2.5 ohms?

Resistors R1 and R2 are connected in parallel, then R12 = R1 * R2 / (R1 + R2) = 100/20 = 5 ohms.

Resistors R12 and R3 are connected in series, then R = R12 + R3 = 5 + 2.5 = 7.5 ohms.

The total resistance of the electrical circuit is equal to: Rob = R + r = 7.5 + 1.5 = 9 Ohm.

Then the current strength in the unbranched circuit is: I = E / Rob = 4.5 / 9 = 0.5 A.

Answer: The current strength is 0.5 A.

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