Question

The current - vortage characteristic of an electric device is as shown in figure (b). The device gets damaged if power dissipated in it exceeds 1 Watt. This device is connected to a dc source of variable emf (V) and a resistance ($R = 100 \Omega$) in series. What is possible range of V for which the device remains operational (i.e. consumes some power) and safe.

Answer 1

20 V < V ≤ 25 V

Answer 2

The problem involves analyzing a series circuit containing a variable DC source, a resistor, and an electric device with given I-V characteristics. We need to find the range of the source voltage (V) for which the device is both operational (consumes some power) and safe (power dissipated does not exceed 1 Watt).

1. Analyze the device's I-V characteristic (Figure b): The current-voltage characteristic shows that the device 'D' starts conducting current only when the voltage across it, $V_D$, reaches 20 V. For $V_D < 20 \text{ V}$, the current $I$ is zero. Once $V_D = 20 \text{ V}$, the device maintains this voltage while allowing current to flow. This indicates it behaves like a voltage regulator or a Zener diode in its breakdown region.

2. Analyze the circuit (Figure a): The circuit consists of a variable EMF source V, a resistor $R = 100 \Omega$, and the device D connected in series. According to Kirchhoff's Voltage Law (KVL) for the series circuit: $V = I \cdot R + V_D$ Substituting the value of R: $V = I \cdot 100 + V_D$

3. Determine the conditions for operational and safe limits:

a) Operational condition: The device is operational if it "consumes some power". Power dissipated in the device is $P_D = V_D \cdot I$. For $P_D > 0$, both $V_D$ and $I$ must be greater than zero. From the I-V characteristic, current $I$ flows only when $V_D = 20 \text{ V}$. Therefore, for the device to be operational, $V_D$ must be exactly 20 V, and the current $I$ must be greater than 0 ($I > 0$). If $V_D = 20 \text{ V}$ and $I > 0$, then from the KVL equation: $V = I \cdot 100 + 20$ Since $I > 0$, it implies $I \cdot 100 > 0$. Thus, $V > 20 \text{ V}$. If $V \le 20 \text{ V}$, then $V_D$ would be less than or equal to 20 V (and $I$ would be 0), so the device would not be operational.

b) Safety condition: The device gets damaged if power dissipated in it exceeds 1 Watt ($P_D > 1 \text{ W}$). So, for safety, $P_D \le 1 \text{ W}$. Since the device is operational, $V_D = 20 \text{ V}$. $P_D = V_D \cdot I = 20 \cdot I$ Applying the safety limit: $20 \cdot I \le 1 \text{ W}$ $I \le \frac{1}{20} \text{ A}$ $I \le 0.05 \text{ A}$

4. Combine the conditions to find the range of V: We have $V_D = 20 \text{ V}$ and $0 < I \le 0.05 \text{ A}$. Substitute $V_D = 20 \text{ V}$ into the KVL equation: $V = I \cdot 100 + 20$

To find the minimum V: As $I$ approaches 0 (but $I > 0$ for operational), $V$ approaches $0 \cdot 100 + 20 = 20 \text{ V}$. So, $V_{min} = 20 \text{ V}$ (exclusive, as I must be greater than 0). This means $V > 20 \text{ V}$.

To find the maximum V: The maximum current allowed is $I_{max} = 0.05 \text{ A}$. Substitute $I_{max}$ into the KVL equation: $V_{max} = (0.05 \text{ A}) \cdot (100 \Omega) + 20 \text{ V}$ $V_{max} = 5 \text{ V} + 20 \text{ V}$ $V_{max} = 25 \text{ V}$ (inclusive, as power is exactly 1 W at this point, which is safe).

Combining both limits, the possible range of V for which the device remains operational and safe is $20 \text{ V} < V \le 25 \text{ V}$.

The final answer is $20 \text{ V} < V \le 25 \text{ V}$

Explanation of the solution: The device operates at a constant voltage of 20 V when conducting. For it to be operational, current must flow, implying V > 20 V. The maximum power dissipation is 1 W, which translates to a maximum current of 0.05 A (since P = VI and V=20V). Using KVL (V = IR + VD), the maximum allowed current (0.05 A) through the 100 Ω resistor gives the maximum source voltage: V_max = (0.05 A)(100 Ω) + 20 V = 25 V. Thus, the safe and operational range for V is from just above 20 V to 25 V.