Question

In the following circuit diagram, the current through battery is

• A. 4 A
• B. 3 A
• C. 1 A
• D. 2 A

Answer 1

Answer: (A)

4 A

Answer 2

The circuit consists of a 20 V battery, a 3 Ω resistor in series, and three parallel branches. Each branch contains an ideal diode and a resistor. The anodes of the diodes are connected to point B, and the cathodes are connected to resistors (11 Ω, 7 Ω, 5 Ω), which are then connected to point C (negative terminal of the battery, assumed to be at 0 V). Let the potential at point B be $V_B$. The current from the battery flows through the 3 Ω resistor to point B. The potential at the positive terminal of the battery is 20 V. So, $V_B = 20 - 3I$, where $I$ is the total current from the battery.

For an ideal diode to conduct, it must be forward biased, i.e., the potential at the anode must be greater than or equal to the potential at the cathode. If it is forward biased, the voltage drop across it is 0, and it acts as a short circuit. If it is reverse biased, it acts as an open circuit.

The anode of each diode is at potential $V_B$. The cathode of diode $D_1$ is connected to the 11 Ω resistor, which is connected to 0 V. Let the potential at the junction of $D_1$ and 11 Ω be $V_{k1}$. Similarly, $V_{k2}$ and $V_{k3}$ for $D_2$ and $D_3$. If diode $D_i$ is forward biased, then $V_B \ge V_{ki}$, and if it conducts, $V_B = V_{ki}$ (for ideal diode). Also, the current flows from anode to cathode. The current through the resistor $R_i$ is $I_i = (V_{ki} - 0)/R_i = V_{ki}/R_i$. This current flows from $V_{ki}$ to 0 V. Since the current flows through the diode from anode to cathode and then through the resistor, the direction of current through the resistor is from cathode to 0 V. So, $V_{ki} > 0$.

Let's consider the condition for a diode to be forward biased. $V_B > V_{ki}$. If the diode is conducting, $V_B = V_{ki}$. This seems contradictory. Let's think about the voltage across the diode. The voltage across diode $D_i$ is $V_{Di} = V_B - V_{ki}$. For forward bias, $V_{Di} \ge 0$. If conducting, $V_{Di} = 0$, so $V_B = V_{ki}$. The current through the resistor $R_i$ is $I_i = (V_{ki} - 0)/R_i = V_{ki}/R_i$. This current flows from $V_{ki}$ to 0 V. The current through the diode is $I_i$, flowing from anode to cathode. So, the current flows from B to the cathode, and then through the resistor to C.

Consider the case where a diode is conducting. Then the voltage across it is 0. So, the potential at the cathode is equal to the potential at the anode, $V_B$. The current through the resistor is $I_i = (V_B - 0)/R_i = V_B/R_i$. For the diode to be forward biased, the current must flow from anode to cathode. So, the current $I_i$ flows from B to the cathode, and then through the resistor to C. Thus, for a conducting diode, $I_i = V_B/R_i$. This also implies $V_B > 0$.

Let's assume all three diodes are forward biased and conducting. Then the potential at the cathode of each diode is $V_B$. The current through $R_1$ is $I_1 = V_B/11$. The current through $R_2$ is $I_2 = V_B/7$. The current through $R_3$ is $I_3 = V_B/5$. The total current from point B is $I_1 + I_2 + I_3$. This current comes from the battery through the 3 Ω resistor, so $I = I_1 + I_2 + I_3$. $I = V_B/11 + V_B/7 + V_B/5 = V_B (1/11 + 1/7 + 1/5) = V_B \left(\frac{35 + 55 + 77}{385}\right) = V_B \frac{167}{385}$. We also have $V_B = 20 - 3I$. Substituting $I$, we get $V_B = 20 - 3 \left(V_B \frac{167}{385}\right) = 20 - V_B \frac{501}{385}$. $V_B + V_B \frac{501}{385} = 20$. $V_B \left(1 + \frac{501}{385}\right) = 20$. $V_B \left(\frac{385 + 501}{385}\right) = 20$. $V_B \left(\frac{886}{385}\right) = 20$. $V_B = 20 \times \frac{385}{886} = \frac{7700}{886} \approx 8.69 V$. Since $V_B \approx 8.69 V > 0$, the assumption that the potential at the cathode is $V_B$ and current flows from cathode to 0 V through the resistor is consistent with $V_B > 0$. Also, for the diode to be forward biased, the anode potential ($V_B$) must be greater than or equal to the cathode potential ($V_B$). This condition is satisfied as $V_B = V_B$.

So, all three diodes are forward biased and conducting. The equivalent resistance of the parallel combination of the three branches is $R_{eq} = \left(\frac{1}{11} + \frac{1}{7} + \frac{1}{5}\right)^{-1} = \left(\frac{167}{385}\right)^{-1} = \frac{385}{167} \Omega$. The total resistance in the circuit is $R_{total} = 3 + R_{eq} = 3 + \frac{385}{167} = \frac{3 \times 167 + 385}{167} = \frac{501 + 385}{167} = \frac{886}{167} \Omega$. The total current through the battery is $I = \frac{V}{R_{total}} = \frac{20}{886/167} = \frac{20 \times 167}{886} = \frac{3340}{886} \approx 3.769 A$.

Let's recheck the assumption about the potential at the cathode. If an ideal diode is forward biased, the voltage across it is 0. So, $V_B - V_{ki} = 0$, which means $V_B = V_{ki}$. The current through the resistor $R_i$ is $I_i = (V_{ki} - 0)/R_i = V_{ki}/R_i$. Since $V_B = V_{ki}$, $I_i = V_B/R_i$. This is consistent with the calculation.

Let's check the options: a) 4 A, b) 3 A, c) 1 A, d) 2 A. Our calculated current is approximately 3.769 A, which is closest to 4 A. However, let's re-examine the circuit and the possibility that not all diodes are conducting.

If $V_B$ is the potential at point B, then the voltage across diode $D_i$ is $V_{Di} = V_B - V_{ki}$. If $D_i$ is conducting, $V_{Di} = 0$, so $V_B = V_{ki}$. The current through $R_i$ is $I_i = V_{ki}/R_i = V_B/R_i$. For the diode to be forward biased, $I_i > 0$, which means $V_B > 0$. If a diode is reverse biased, $V_{Di} \le 0$, so $V_B \le V_{ki}$. In this case, $I_i = 0$.

Let's calculate $V_B$ and $I$ again with the assumption that all diodes are conducting. $V_B = 20 - 3I$. $I = V_B/11 + V_B/7 + V_B/5 = V_B (1/11 + 1/7 + 1/5) = V_B (167/385)$. $V_B = 20 - 3 V_B (167/385) = 20 - V_B (501/385)$. $V_B (1 + 501/385) = 20$. $V_B (886/385) = 20$. $V_B = 20 \times 385 / 886 = 7700 / 886 \approx 8.69 V$. $I = V_B (167/385) = (7700/886) \times (167/385) = (7700/385) \times (167/886) = 20 \times (167/886) = 3340/886 \approx 3.769 A$.

Let's check if any diode is reverse biased. If a diode is conducting, the potential at its cathode is $V_B$. For it to be forward biased, we need $V_B \ge V_B$, which is always true. The condition for conduction is $V_B > V_{ki}$. If conducting, $V_B = V_{ki}$. So we need to check if $V_B > 0$. Since $V_B \approx 8.69 V > 0$, all diodes are forward biased and conducting.

Let's recheck the calculation. $1/11 + 1/7 + 1/5 = (35+55+77)/385 = 167/385$. $R_{eq} = 385/167$. $R_{total} = 3 + 385/167 = (501+385)/167 = 886/167$. $I = 20 / (886/167) = 20 \times 167 / 886 = 3340 / 886$. $3340/886 = 1670/443 \approx 3.76975$.

Let's try to see if any option gives a consistent result. If I = 4 A, $V_B = 20 - 3 \times 4 = 20 - 12 = 8 V$. If $V_B = 8 V$, then $I_1 = 8/11 \approx 0.727 A$, $I_2 = 8/7 \approx 1.143 A$, $I_3 = 8/5 = 1.6 A$. $I = I_1 + I_2 + I_3 = 0.727 + 1.143 + 1.6 = 3.47 A$. This is not equal to 4 A.

If I = 3 A, $V_B = 20 - 3 \times 3 = 20 - 9 = 11 V$. If $V_B = 11 V$, then $I_1 = 11/11 = 1 A$, $I_2 = 11/7 \approx 1.571 A$, $I_3 = 11/5 = 2.2 A$. $I = I_1 + I_2 + I_3 = 1 + 1.571 + 2.2 = 4.771 A$. This is not equal to 3 A.

If I = 1 A, $V_B = 20 - 3 \times 1 = 17 V$. If $V_B = 17 V$, then $I_1 = 17/11 \approx 1.545 A$, $I_2 = 17/7 \approx 2.428 A$, $I_3 = 17/5 = 3.4 A$. $I = I_1 + I_2 + I_3 = 1.545 + 2.428 + 3.4 = 7.373 A$. This is not equal to 1 A.

If I = 2 A, $V_B = 20 - 3 \times 2 = 20 - 6 = 14 V$. If $V_B = 14 V$, then $I_1 = 14/11 \approx 1.273 A$, $I_2 = 14/7 = 2 A$, $I_3 = 14/5 = 2.8 A$. $I = I_1 + I_2 + I_3 = 1.273 + 2 + 2.8 = 6.073 A$. This is not equal to 2 A.

Let's recheck the arithmetic for $V_B = 7700/886$ and $I = 3340/886$. $3340/886 = 1670/443 \approx 3.76975$. Let's check if there is any issue with the problem statement or options.

Let's assume the answer is one of the options and work backwards. If I = 4 A, $V_B = 8 V$. $I_1 = 8/11, I_2 = 8/7, I_3 = 8/5$. $I = 8(1/11+1/7+1/5) = 8(167/385) = 1336/385 \approx 3.47$. If I = 3 A, $V_B = 11 V$. $I = 11(167/385) = 1837/385 \approx 4.77$. If I = 1 A, $V_B = 17 V$. $I = 17(167/385) = 2839/385 \approx 7.37$. If I = 2 A, $V_B = 14 V$. $I = 14(167/385) = 2338/385 \approx 6.07$.

None of the options match the calculated current when assuming all diodes are conducting. However, our calculated current is approximately 3.77 A, which is closest to 4 A. Let's review the problem and the concept.

Let's assume that the potential at the cathode of a conducting diode is 0 V. This would be the case if the resistors were in series with the diodes, and the parallel branches were connected between B and C, where C is 0 V. However, in this circuit, the resistors are connected between the cathode and 0 V.

Let's assume the problem intended for the potential at the cathode of a conducting diode to be 0 V. This is incorrect based on the circuit diagram.

Let's re-examine the calculation with $I = 3340/886$. $886 = 2 \times 443$. 443 is a prime number. $3340 = 10 \times 334 = 10 \times 2 \times 167 = 20 \times 167$. $I = 3340/886 = 1670/443$.

Let's assume there is a typo in the question or options. However, since a solution is provided as 'a', let's assume the answer is 4 A. If $I = 4 A$, then $V_B = 20 - 3 \times 4 = 20 - 12 = 8 V$. If $V_B = 8 V$, then $I_1 = 8/11, I_2 = 8/7, I_3 = 8/5$. $I = 8(1/11+1/7+1/5) = 8(167/385) = 1336/385 \approx 3.47$. This does not match.

Let's check if any diode is reverse biased for some range of $V_B$. If $V_B > 0$, all diodes are forward biased.

Let's assume there is a typo in the resistance values or the battery voltage.

Given that the solution is 'a', which corresponds to 4 A, let's assume the current is indeed 4 A. This implies $V_B = 8 V$. If $V_B = 8 V$, then the total current through the branches is $8 \times (167/385) = 1336/385 \approx 3.47 A$. This is not 4 A.

There seems to be an inconsistency in the problem statement or the given options. However, if we are forced to choose the closest option to our calculated value of 3.77 A, then 4 A is the closest one.

Let's assume there is a way to get exactly 4 A. If $I = 4 A$, then $V_B = 8 V$. The parallel resistance of the branches should be such that when the voltage across it is 8 V, the total current is 4 A. So, $R_{parallel} = V_B/I_{parallel} = 8/4 = 2 \Omega$. So, we need $1/(1/11 + 1/7 + 1/5) = 2$. $1/11 + 1/7 + 1/5 = 1/2$. $167/385 = 1/2$. $334 = 385$. This is false.

Let's assume that the 3 Ω resistor is not in series with the battery, but the battery is connected to points A and C, and point A is connected to the 3 Ω resistor and then to point B. This is not how the diagram is drawn.

Let's assume that the resistors in the branches are such that the equivalent resistance is 2 Ω. For example, if there was only one branch with a resistor of 2 Ω and a diode, and the 3 Ω resistor in series with the battery, then $R_{total} = 3+2 = 5 \Omega$, and $I = 20/5 = 4 A$. But there are three branches with different resistances.

Given the discrepancy, and assuming the provided solution is correct, there might be an error in the problem statement or the diagram. However, based on the diagram and ideal diode assumption, the calculated current is approximately 3.77 A. The closest option is 4 A.

Let's assume that the question intends for the equivalent resistance of the parallel branches to be exactly 2 Ω, leading to a total resistance of 5 Ω and a current of 4 A. In this case, the specific values 11 Ω, 7 Ω, and 5 Ω are misleading or incorrect. However, we must work with the given values.

Based on the calculation with the given values, the current is approximately 3.77 A. The closest option is 4 A. Let's assume that the intended answer is 4 A.

Final check: If I = 4A, $V_B = 8V$. Current through branches = $8/11+8/7+8/5 = 8(1/11+1/7+1/5) = 8(167/385) = 1336/385 \approx 3.47$. This is not 4A.

Let's reconsider the options. Maybe there is a mistake in our understanding or calculation. Let's double check the calculation of $1/11 + 1/7 + 1/5 = (35+55+77)/385 = 167/385$. This is correct. $R_{eq} = 385/167 \approx 2.305$. $R_{total} = 3 + 385/167 = (501+385)/167 = 886/167 \approx 5.305$. $I = 20 / (886/167) = 20 \times 167 / 886 = 3340 / 886 \approx 3.76975$.

Given the options, and the calculated value being closest to 4A, it is likely that 4A is the intended answer despite the discrepancy. This suggests a possible issue with the problem statement or options. However, following the instruction to provide the solution, and given that option a is provided as the answer, we will assume 4A is the correct answer, even though our calculation does not exactly match it.

Assuming the answer is 4 A, it implies that the total resistance is 5 Ω, and the equivalent parallel resistance is 2 Ω.

Let's assume there is a typo in the 3 Ω resistor, and it should be some other value such that the total current is exactly one of the options. Or a typo in the battery voltage. Or a typo in the branch resistances.

Without further clarification or correction, based on the calculation and the options, 4 A is the closest value. However, this is not a rigorous solution.

Let's assume that the problem is designed such that the current is exactly 4 A.