Question

A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose resistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter?

Answer 1

Answer on Question 71070, Physics, Other

Question:

A 2 Ω resistor is connected in a series with a 20.0 V battery and a three-branch parallel network with branches whose resistance are 8 Ω each. Ignoring the battery's internal resistance, what is the current in the battery?

Solution:

Let's first find the equivalent resistance of the three-branch parallel network:

$$\frac {1}{R _ {e q p a r a l l e l}} = \frac {1}{R _ {2}} + \frac {1}{R _ {3}} + \frac {1}{R _ {4}} = \frac {1}{8 \Omega} + \frac {1}{8 \Omega} + \frac {1}{8 \Omega} = \frac {3}{8} \Omega .$$ $$R _ {e q p a r a l l e l} = \frac {8}{3} \Omega .$$

This three-branch parallel network of resistors is connected in series with the 2.0-ohm resistor. So, the equivalent resistance of the circuit is equal:

$$R _ {e q} = R _ {1} + R _ {e q p a r a l l e l} = 2 \Omega + \frac {8}{3} \Omega = \frac {14}{3} \Omega .$$

Finally, we can find the current in the battery from the Ohm's law:

$$I = \frac {V}{R _ {e q}} = \frac {20 V}{\frac {14}{3} \Omega} = 4.28 \text{ A}.$$

Answer:

$$I = 4.28 \text{ A}.$$

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