Q.An electrical device draws $2\ \text{kW}$ power from AC mains (voltage $223\ \text{V}$ (rms) $= \sqrt{50000}\ \text{V}$). The current differs (lags) in phase by $\phi$ $\left(\tan\phi = -\dfrac{3}{4}\right)$ as compared to voltage. Find

Concept and Intuition

The power consumed by a device at a fixed voltage is given by $P = \frac{V^2}{R}$.

If you change the resistance, the power changes inversely.

Here, cutting a coil into two equal parts and then connecting them in parallel dramatically lowers the total resistance, so the power (energy per second) skyrockets.

The trick is to track how the resistance changes step by step.

Step-by-step reasoning

1. Original coil Rated $400 \, \text{W}$ at $200 \, \text{V}$. Using $P = \frac{V^2}{R}$, we find the original resistance:

2. Cut into two equal parts Each part is half the length of the original wire. Since resistance is proportional to length (for uniform wire), each half has resistance:

3. Connect the two halves in parallel For two equal resistors $R_{\text{half}}$ in parallel, the equivalent resistance is:

4. New power (energy per second) With the same source voltage $200 \, \text{V}$, the new power is:

5. Percentage increase Increase in power = $1600 - 400 = 1200 \, \text{W}$. Percentage increase = $\frac{1200}{400} \times 100\% = 300\%$. …

Step 1 — Write the power relation for a resistive AC load

Given $P = 100\ \text{W}$ and $V_{\text{rms}} = 220\ \text{V}$ (the 50 Hz frequency does not affect a purely resistive load's current): …

Step 1 — Total line resistance

The substation is 20 km away, and it is a two-wire line, so the total wire length is

Step 2 — Line current

Power is transmitted from the generating station at 45000 V (the stepdown to 220 V happens only at the town's substation), so the current drawn is …

Concept & Intuition

The circuit has two parallel branches: one contains only the 16 Ω resistor; the other contains a 6 Ω and a 4 Ω resistor in series. The same voltage appears across both branches because they share the same two nodes. If we know the power in the 16 Ω resistor, we can find the current through it, then the voltage across it, which is also the voltage across the series combination. From that voltage we can find the current in the series branch, and finally the power in the 4 Ω resistor.

Step‑by‑step reasoning

1. Find the current through the 16 Ω resistor. Power in a resistor is $P = I^2 R$. For the 16 Ω resistor, $P_{16} = 2\ \text{W}$.

2. Find the voltage across the 16 Ω resistor (and hence across the other branch). Using Ohm’s law: $V = I R$.

This voltage $V = 4\sqrt{2}\ \text{V}$ is the same across the series combination of 6 Ω and 4 Ω.

3. Find the current through the series branch (6 Ω + 4 Ω). The total resistance in that branch is $R_{\text{series}} = 6 + 4 = 10\ \Omega$. Using Ohm’s law:

This current flows through both the 6 Ω and the 4 Ω resistor. …

The key concept is Ohm’s law for AC circuits with pure resistance. In a purely resistive circuit, voltage and current are in phase, and the relationship between instantaneous values is exactly the same as for DC: $i(t) = \frac{v(t)}{R}$. There is no reactance, no phase shift, and no RMS conversion needed when the question asks for peak current.

1. Identify the given voltage function The voltage is $V(t) = 400 \sin(\pi t)$. The amplitude (peak value) of this sinusoidal voltage is the coefficient in front of the sine function:

2. Apply Ohm’s law for peak values For a resistor, the peak current is directly given by:

This is because the resistor obeys $V = IR$ at every instant, so the maximum of current occurs exactly when voltage is maximum.

3. Plug in the numbers

4. Match to the options The value 1.42857 A rounds to 1.43 A, which is option (D). …

The key concept here is that for a device like a bulb (which behaves approximately as a fixed resistor), power is proportional to the square of the voltage: $P = \frac{V^2}{R}$. When the voltage changes by a small percentage, the power changes by roughly twice that percentage, because squaring amplifies the relative change. This is a classic application of differential approximation or simple algebra.

Let’s work through it step by step.

1. Identify the relationship. The bulb is rated at $220\,\text{V}$ and $50\,\text{W}$. Assuming its resistance $R$ is constant (a reasonable approximation for a filament bulb at steady temperature), we have:

So power is proportional to the square of the applied voltage.

2. Express the voltage drop. The voltage drops by $5\%$ of its rated value. That means the new voltage is:

3. Find the new power. Since $P \propto V^2$, the new power is:

So the new power is $90.25\%$ of the rated power.

4. Compute the percentage decrease. The decrease in power is:

This is very close to $10\%$.

5. Why not exactly 10%? …

Concept and Intuition

When bulbs are rated with a power (like 40 W, 60 W, 100 W), that rating is the power they would consume if connected directly to the rated voltage (here, presumably 220 V). From $P = V^2 / R$, we see that for a fixed voltage, lower wattage means higher resistance. So the 40 W bulb has the highest resistance, and the 100 W bulb the lowest.

Now, when these bulbs are connected in series, the same current flows through each. Power dissipated in a resistor is $I^2 R$. Since current is identical, the bulb with the largest resistance dissipates the most power and therefore glows the brightest.

Step-by-step reasoning

1. Find the resistance of each bulb from its rating. For a bulb rated $P$ at voltage $V$, we have $R = V^2 / P$. Assuming the rated voltage is 220 V (the source voltage),

Clearly, $R_{40} > R_{60} > R_{100}$.

2. In a series circuit, current is the same through all bulbs.

   Total resistance $R_{\text{total}} = R_{40} + R_{60} + R_{100}$.

   Current $I = \frac{220}{R_{\text{total}}}$.

3. Power dissipated in each bulb is $P_{\text{actual}} = I^2 R$. …

Concept & Intuition

A heating unit is essentially a resistor. For a purely resistive load, the power dissipated (heat output) is given by $P = V^2 / R$, where $R$ is constant (the heater’s resistance doesn’t change significantly with a modest voltage drop). This means power is proportional to the square of the voltage. So if the voltage drops, the power drops much faster — a classic quadratic relationship. The problem asks for the percentage drop in heat output, not the remaining power.

Step-by-step solution

1. Find the heater’s resistance from the design conditions. At the designed voltage $V_1 = 400\ \mathrm{V}$ and power $P_1 = 500\ \mathrm{W}$:

This resistance is fixed.

2. Compute the new power when the voltage drops to $V_2 = 160\ \mathrm{V}$. Using the same resistance:

3. Find the drop in power (absolute):

4. Express the drop as a percentage of the original power: $\text{Percentage drop} = \frac{\Delta P}{P_1} \times 100\% = \frac{420}{500} \times 100\% = 84\%.$ …

Why only the resistor?

The key idea is the difference between dissipative and reactive components. A resistor converts electrical energy into heat irreversibly — that's power dissipation. Inductors and capacitors, on the other hand, store energy in magnetic and electric fields respectively, and return that energy to the circuit each cycle. Over a full AC cycle, the net power absorbed by a pure inductor or a pure capacitor is zero.

This is why in AC circuit analysis, we talk about real power (dissipated in R, measured in watts) and reactive power (shuttled between L and C, measured in VAR). Only R contributes to the real power.

Step-by-step reasoning

1. Power dissipation in a resistor

   For a resistor $R$ carrying current $i$, the instantaneous power is $p = i^2 R$, which is always positive (or zero). Over any time interval, the average power is $\langle P \rangle = I_{\text{rms}}^2 R$, which is non-zero. This energy leaves the circuit as heat.

2. Power in an ideal inductor

   For an inductor $L$, the voltage-current relation is $v = L \frac{di}{dt}$. The instantaneous power is $p = v i = L i \frac{di}{dt}$. Over a full AC cycle, the energy stored in the magnetic field ($\frac{1}{2} L i^2$) goes up and then back down to zero — the integral of $p$ over one cycle is exactly zero. No net energy is lost.

3. Power in an ideal capacitor

   For a capacitor $C$, $i = C \frac{dv}{dt}$, so $p = v i = C v \frac{dv}{dt}$. Similarly, the energy stored in the electric field ($\frac{1}{2} C v^2$) is returned to the circuit each cycle. Net power dissipation is zero.

4. In a series LCR circuit …