Question 455: Current Electricity

Question

In the following circuit, current through ACB if each resistance $\mathrm{R}=4 \Omega$ is

Accepted Answer

The total voltage across A and B is 16 V . Since the bridge is balanced, the circuit can be simplified by considering the parallel combination of the two paths:
Path ACB: $\mathrm{R}_{\mathrm{AC}}+\mathrm{R}_{\mathrm{CB}}=4 \Omega+4 \Omega=8 \Omega$
Path ADB: $\mathrm{R}_{\mathrm{AD}}+\mathrm{R}_{\mathrm{DB}}=4 \Omega+4 \Omega=8 \Omega$
The equivalent resistance between A and B is:
$$ \mathrm{R}_{\mathrm{eq}}=\frac{8 	imes 8}{8+8}=\frac{64}{16}=4 \Omega $$
The total current from the 16 V battery is:
$$ I_{	ext {total }}=\frac{V}{R_{e q}}=\frac{16}{4}=4 A $$
This total current of 4A splits equally between the two parallel paths ACB and ADB because the resistances are equal:
$$ I_{A C B}=\frac{I_{	ext {total }}}{2}=\frac{4}{2}=2 \mathrm{~A} $$

Answer

1 A

Answer

2 A

Answer

3 A

Answer

4 A

Current Electricity

In the following circuit, current through ACB if each resistance \mathrm{R}=4 \Omega is

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