The current passing through the battery in the given circuit, is:
1. 2.0 A
2. 0.5 A
3. 2.5 A
4. 1.5 A
Best answer
Answer: 2. 0.5 A
Key idea: the middle network is a balanced Wheatstone bridge, so the \(6\ \Omega\) resistor carries no current.
Points C and D are joined by a wire, and A and F are joined by a wire. Between B and E the bridge arms are: \(B\to A=5\ \Omega\), \(B\to D=2.5\ \Omega\), \(A\to E=3\ \Omega\), \(D\to E=1.5\ \Omega\), with \(6\ \Omega\) between A and D.
Since \(\dfrac{5}{2.5}=\dfrac{3}{1.5}\), the bridge is balanced and the \(6\ \Omega\) can be ignored. Then \((5+3)=8\ \Omega\) is in parallel with \((2.5+1.5)=4\ \Omega\):
$$R_{BE}=\frac{8\times4}{12}=\frac83\ \Omega$$Total resistance \(=1.5+5.5+\tfrac13+\tfrac83=10\ \Omega\).
$$I=\frac{5}{10}=0.5\ \mathrm{A}$$Log in or create a free account to answer this question.
