Step: Find the equivalent resistance between \(A\) and \(D.\)
The circuit can be redrawn as shown in the figure below.
As we can see, there will be no current in the resistor between \(B\) and \(C,\) it is short-circuited.
The equivalent resistance between \(A\) and \(D\) is given by;
\(\frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{2}+1\)
\(\Rightarrow R_{eq}=\frac{1}{2}~\Omega \)
Hence, option (3) is the correct answer.
Two cities are \(150~\text{km}\) apart. The electric power is sent from one city to another city through copper wires. The fall of potential per km is \(8~\text{volts}\) and the average resistance per \(\text{km}\) is \(0.5~\text{ohm}.\) The power loss in the wire is:
Hint: \(P=\dfrac{V_{\text {total }}^2}{R_{\text {total }}}\)
Step: Find the power loss in the wire.
The total potential drop \((V_{total})\) is given by;
\(\Rightarrow V_{\text {total }}=\text { Potential drop per km } \times \text { Distance in km} \)
\(\Rightarrow V_{\text {total }}=8 ~\text{V}/ \text{km} \times 150 ~\text{km}=1200~\text{V}\)
The total resistance \((R_{total})\) over the distance is given by;
\(\Rightarrow R_{\text {total }}=\text { Resistance per km} \times \text { Distance in km} \)
\(\Rightarrow R_{\text {total }}=0.5~ \Omega / \text{km} \times 150 ~\text{km}=75 ~\Omega \)
The power loss \((P)\) in the wire is given by;
\(\Rightarrow P=\frac{V_{\text {total }}^2}{R_{\text {total }}}\)
\(\Rightarrow P=\frac{(1200 ~\text{V})^2}{75~\Omega} \Rightarrow P=19.2~\text{kW}\)
Hence, option (2) is the correct answer.