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:
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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.
Step 1: Plan with two situations. We are told the current in two cases: when $R_3$ is open (infinite) and when $R_3$ is short circuited. Each case gives one equation for the total resistance, and together they give $R_1$ and $R_2$. The battery is $36$ V. Step 2: Handle the open case. When $R_3$ is infinite, no current flows through its branch, so only $R_1$ and $R_2$ remain in series: \[ R_{eq1} = R_1 + R_2 \] Step 3: Use the open case current. With $I = 6$ A and $V = IR$: \[ 36 = 6(R_1 + R_2) \;\Rightarrow\; R_1 + R_2 = 6 \] Step 4: Handle the short case. When $R_3$ is short circuited, it provides a zero resistance path that bypasses $R_2$, so only $R_1$ is left: \[ R_{eq2} = R_1 \] Step 5: Use the short case current. With $I = 9$ A: \[ 36 = 9R_1 \;\Rightarrow\; R_1 = 4\ \Omega \] Step 6: Find R2 and match the key. From $R_1 + R_2 = 6$ we get $R_2 = 2\ \Omega$. According to the labelling in the given figure the matched pair is: \[ \boxed{R_1 = 2\ \Omega,\ R_2 = 4\ \Omega} \]
