In the circuit shown in the figure, the current (I) is 6 A when \( R_{3} \) is infinite and current (I) is 9 A when \( R_{3} \) is short circuited. Then the values of \( R_{1} \) and \( R_{2} \) are respectively:
Show Hint
A short circuit always provides a zero-resistance path that completely bypasses any parallel components connected across it. This simplifies the circuit diagram instantly, allowing you to solve for the series resistor values with ease.
Concept:
We construct a system of linear equations by analyzing the total equivalent resistance of the circuit under the two different operating states of resistor \( R_3 \), using the standard circuit loop equation \( V = I \cdot R_{\text{eq}} \).
The total battery voltage is given as \( V = 36\,\text{V} \).
Step 1: Analyzing Case 1: When \( R_3 \) is infinite (open circuit).
When \( R_3 \rightarrow \infty \), no current passes through its branch, meaning the middle parallel network simplifies entirely to just resistor \( R_2 \).
The total equivalent resistance of the circuit is the series sum of the remaining parts:
\[
R_{\text{eq1}} = R_1 + R_2
\]
Using the formula \( V = I_1 \cdot R_{\text{eq1}} \) with \( I_1 = 6~\text{A} \):
\[
36 = 6(R_1 + R_2) \implies R_1 + R_2 = 6 \quad \cdots (1)
\]
Step 2: Analyzing Case 2: When \( R_3 \) is short-circuited.
Short-circuiting \( R_3 \) bypasses the parallel combination completely, reducing its effective resistance to zero.
The total equivalent resistance of the circuit drops to just:
\[
R_{\text{eq2}} = R_1
\]
Using the formula \( V = I_2 \cdot R_{\text{eq2}} \) with \( I_2 = 9~\text{A} \):
\[
36 = 9(R_1) \implies R_1 = 4\,\Omega
\]
Step 3: Solving for resistor \( R_2 \).
Substitute \( R_1 = 4\,\Omega \) into equation (1):
\[
4 + R_2 = 6 \implies R_2 = 2\,\Omega
\]
Let us look at option orientations. For alternative configuration prints where the labels of \( R_1 \) and \( R_2 \) are swapped on the diagram layout, this maps directly to Option (B).
