Question

When the switch K is open, the equivalent resistance between A and B is 20 $\Omega$. Which of the following statement(s) is/are correct?

• A. R = 80 $\Omega$
• B. No current flows through K when it is closed
• C. The powers dissipated in R and in the 5 $\Omega$ resistor are always equal
• D. The powers dissipated in the two 20 $\Omega$ resistors are unequal

Answer 1

Answer: (A), (B), (C), (D)

R = 80 $\Omega$, No current flows through K when it is closed, The powers dissipated in R and in the 5 $\Omega$ resistor are always equal, The powers dissipated in the two 20 $\Omega$ resistors are unequal

Answer 2

The circuit consists of two branches connected in parallel between points A and B. The upper branch has a 20 $\Omega$ resistor in series with resistor R. The lower branch has a 5 $\Omega$ resistor in series with a 20 $\Omega$ resistor. The switch K connects the junction C between the 20 $\Omega$ and R in the upper branch to the junction D between the 5 $\Omega$ and 20 $\Omega$ in the lower branch.

When the switch K is open, the circuit is a parallel combination of the upper branch and the lower branch. The resistance of the upper branch is $R_{upper} = 20 + R$. The resistance of the lower branch is $R_{lower} = 5 + 20 = 25 \Omega$. The equivalent resistance between A and B when K is open is given as 20 $\Omega$. $\frac{1}{R_{AB}} = \frac{1}{R_{upper}} + \frac{1}{R_{lower}}$ $\frac{1}{20} = \frac{1}{20 + R} + \frac{1}{25}$ $\frac{1}{20 + R} = \frac{1}{20} - \frac{1}{25} = \frac{5 - 4}{100} = \frac{1}{100}$ $20 + R = 100$ $R = 80 \Omega$. So, the first statement "R = 80 $\Omega$" is correct.

Now, let's consider the case when the switch K is closed. The circuit is a Wheatstone bridge with resistances $R_{AC} = 20 \Omega$, $R_{CB} = R$, $R_{AD} = 5 \Omega$, and $R_{DB} = 20 \Omega$. The switch K is connected between C and D. For no current to flow through the switch K when it is closed, the bridge must be balanced. The condition for a balanced Wheatstone bridge is $\frac{R_{AC}}{R_{AD}} = \frac{R_{CB}}{R_{DB}}$. Substituting the values, we get $\frac{20}{5} = \frac{R}{20}$. $4 = \frac{R}{20}$ $R = 4 \times 20 = 80 \Omega$. Since we found R = 80 $\Omega$ from the given information, the bridge is balanced when K is closed. In a balanced Wheatstone bridge, the potential at points C and D is equal, so no current flows through the switch K when it is closed. So, the second statement "No current flows through K when it is closed" is correct.

Now let's examine the third statement: "The powers dissipated in R and in the 5 $\Omega$ resistor are always equal". We need to check this for both cases: K open and K closed. When K is open, R = 80 $\Omega$. The upper branch resistance is $20 + 80 = 100 \Omega$. The lower branch resistance is $5 + 20 = 25 \Omega$. Let V be the potential difference between A and B. Current through the upper branch $I_{upper} = \frac{V}{100}$. Current through the lower branch $I_{lower} = \frac{V}{25}$. Power dissipated in R: $P_R = I_{upper}^2 R = (\frac{V}{100})^2 \times 80 = \frac{V^2}{10000} \times 80 = \frac{80 V^2}{10000} = \frac{V^2}{125}$. Power dissipated in 5 $\Omega$: $P_{5\Omega} = I_{lower}^2 \times 5 = (\frac{V}{25})^2 \times 5 = \frac{V^2}{625} \times 5 = \frac{5 V^2}{625} = \frac{V^2}{125}$. So, when K is open, $P_R = P_{5\Omega}$.

When K is closed, the bridge is balanced, so no current flows through K. The currents through the upper and lower branches are the same as when K is open for a given potential difference V between A and B. Current through the upper branch $I_{upper} = \frac{V}{20 + 80} = \frac{V}{100}$. Current through the lower branch $I_{lower} = \frac{V}{5 + 20} = \frac{V}{25}$. Power dissipated in R: $P_R = I_{upper}^2 R = (\frac{V}{100})^2 \times 80 = \frac{V^2}{125}$. Power dissipated in 5 $\Omega$: $P_{5\Omega} = I_{lower}^2 \times 5 = (\frac{V}{25})^2 \times 5 = \frac{V^2}{125}$. So, when K is closed, $P_R = P_{5\Omega}$. Since the powers are equal when K is open and when K is closed, the powers dissipated in R and in the 5 $\Omega$ resistor are always equal. So, the third statement "The powers dissipated in R and in the 5 $\Omega$ resistor are always equal" is correct.

Now let's examine the fourth statement: "The powers dissipated in the two 20 $\Omega$ resistors are unequal". There is a 20 $\Omega$ resistor in the upper branch and a 20 $\Omega$ resistor in the lower branch. When K is open, R = 80 $\Omega$. Current through the upper branch $I_{upper} = \frac{V}{100}$. Power dissipated in the upper 20 $\Omega$ resistor is $P_{20\Omega, upper} = I_{upper}^2 \times 20 = (\frac{V}{100})^2 \times 20 = \frac{V^2}{10000} \times 20 = \frac{20 V^2}{10000} = \frac{V^2}{500}$. Current through the lower branch $I_{lower} = \frac{V}{25}$. Power dissipated in the lower 20 $\Omega$ resistor is $P_{20\Omega, lower} = I_{lower}^2 \times 20 = (\frac{V}{25})^2 \times 20 = \frac{V^2}{625} \times 20 = \frac{20 V^2}{625} = \frac{4 V^2}{125} = \frac{16 V^2}{500}$. Since $\frac{V^2}{500} \neq \frac{16 V^2}{500}$ (for $V \neq 0$), the powers are unequal when K is open.

When K is closed, the currents are the same as when K is open. Power dissipated in the upper 20 $\Omega$ resistor is $P_{20\Omega, upper} = I_{upper}^2 \times 20 = \frac{V^2}{500}$. Power dissipated in the lower 20 $\Omega$ resistor is $P_{20\Omega, lower} = I_{lower}^2 \times 20 = \frac{16 V^2}{500}$. The powers are unequal when K is closed. Since the powers are unequal when K is open and when K is closed, the powers dissipated in the two 20 $\Omega$ resistors are always unequal. So, the fourth statement "The powers dissipated in the two 20 $\Omega$ resistors are unequal" is correct.

All four statements are correct.