Question

Two resistances are connected in parallel. Equivalent resistance of combination is \(\tfrac{6}{5}\ \Omega\). When one resistance is broken out, resistance becomes \(2\ \Omega\). The resistance of broken wire was:

• A. \(\tfrac{2}{5}\ \Omega\)
• B. \(2\ \Omega\)
• C. \(3\ \Omega\)
• D. \(\tfrac{5}{2}\ \Omega\)

Hint:

The remaining \(2\ \Omega\) is one resistor; put it into \(\frac{R_1R_2}{R_1+R_2}=\frac{6}{5}\) and solve.

Answer 1

The Correct Option is C

Step 1: Identify the surviving resistance.
When one wire breaks, only the other resistor remains and the circuit resistance is that resistor's value. So one resistance is \(R_1 = 2\ \Omega\).
Step 2: Write the parallel-combination formula.
\[ \frac{R_1 R_2}{R_1 + R_2} = \frac{6}{5} \]
Step 3: Substitute \(R_1 = 2\ \Omega\).
\[ \frac{2 R_2}{2 + R_2} = \frac{6}{5} \]
Step 4: Solve for \(R_2\) (cross-multiply).
\[ 5 (2 R_2) = 6 (2 + R_2) \]
\[ 10 R_2 = 12 + 6 R_2 \]
\[ 4 R_2 = 12 \Rightarrow R_2 = 3\ \Omega \]
Step 5: The broken wire is the one that is no longer in circuit, i.e. \(R_2 = 3\ \Omega\).
\[\boxed{R_{\text{broken}} = 3\ \Omega}\]