Home › Physics › Q
PhysicsCurrent Electricity
In the circuit shown, $L = 2$ H, $R = 5\ \Omega$. At an instant when current $i = 2$ A and rate of change of current $di/dt = 1$ A/s, the potential difference $V_A - V_B$ is:
Options
1
$10$ V
2
$9$ V
3
$12$ V
4
$14$ V
Correct Answer
$12$ V
Solution
1

For a series RL circuit from A to B:

$$V_A - V_B = iR + L\frac{di}{dt}$$
2
$$= (2)(5) + (2)(1) = 10 + 2 = \boxed{12 \text{ V}}$$
$V_A - V_B = iR + L\frac{di}{dt} = 10 + 2 = 12$ V
Theory: Current Electricity
1. Inductors in Circuits

Self-inductance $L$: $\mathcal{E} = -L(di/dt)$. The inductor opposes change in current (back-EMF). Voltage across inductor: $V_L = L(di/dt)$. If current is increasing ($di/dt > 0$): potential drops in direction of current flow. Total voltage across RL series: $V = iR + L(di/dt)$. Energy stored in inductor: $U = \frac{1}{2}Li^2$. Inductance of solenoid: $L = \mu_0 n^2 V$ where $V$ = volume.

2. RL Circuit Transient

When DC source $\mathcal{E}$ connected to RL: $i(t) = \frac{\mathcal{E}}{R}(1 - e^{-t/\tau})$ where $\tau = L/R$ = time constant. At $t=0$: $i=0$ (inductor opposes sudden change). At $t=\infty$: $i = \mathcal{E}/R$ (steady state, inductor acts as wire). At $t = \tau$: $i = 0.632 \mathcal{E}/R$. When source disconnected: $i(t) = I_0 e^{-t/\tau}$ (current decays). Inductor tries to maintain current.

3. Faradays Law of Induction

$\mathcal{E} = -d\Phi_B/dt$ (magnitude: $\mathcal{E} = d\Phi/dt$). Lenz law: induced current opposes change causing it (negative sign). Motional EMF: $\mathcal{E} = Bvl$ (conductor of length $l$ moving at $v$ in field $B$). Eddy currents: induced in bulk conductors. Used in: induction cooking, electromagnetic braking, metal detectors. Reduced by lamination in transformers.

4. Mutual Inductance

$M$: $\mathcal{E}_2 = -M(di_1/dt)$. EMF in coil 2 due to changing current in coil 1. $M = \mu_0 N_1 N_2 A/l$ for coaxial solenoids. Coefficient of coupling: $k = M/\sqrt{L_1 L_2}$. Transformer: $V_s/V_p = N_s/N_p$. $I_s/I_p = N_p/N_s$ (ideal). Energy: $U = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2$.

5. LC Oscillations

LC circuit: charge oscillates. $q = Q_0\cos(\omega t)$ where $\omega = 1/\sqrt{LC}$. Frequency $f = 1/(2\pi\sqrt{LC})$. Energy transfers between $E = Q^2/(2C)$ (capacitor) and $B = LI^2/2$ (inductor). Total energy constant (no resistance). Analogous to SHM: $q \leftrightarrow x$, $L \leftrightarrow m$, $1/C \leftrightarrow k$. Radio tuning uses variable $C$ to change $f$.

6. Kirchhoffs Laws

KCL: $\sum I_{in} = \sum I_{out}$ at junction. KVL: $\sum \mathcal{E} = \sum IR$ around loop. Sign convention: EMF positive if traversed from $-$ to $+$ terminal. Voltage drop across $R$ positive if traversed in direction of current. For inductor: voltage drop $= L(di/dt)$ if traversed in current direction. Apply KVL to solve complex circuits with multiple loops.

7. Magnetic Flux and Gausss Law for B

$\Phi_B = \int \vec{B}\cdot d\vec{A} = BA\cos\theta$ (uniform field). Gauss law for magnetism: $\oint \vec{B}\cdot d\vec{A} = 0$ (no magnetic monopoles). Field lines form closed loops (unlike electric field lines which start/end on charges). Total flux through any closed surface = 0. Flux linkage: $\Lambda = N\Phi = Li$ for solenoid.

8. Energy in Magnetic Field

Energy density in magnetic field: $u_B = B^2/(2\mu_0)$ (J/m³). Total energy stored in solenoid (field inside): $U = u_B \times V = \frac{B^2}{2\mu_0}Al = \frac{1}{2}Li^2$. Compare electric field energy density: $u_E = \epsilon_0 E^2/2$. In EM wave: $u_E = u_B$ (equal energy in $E$ and $B$ fields). Energy flows via Poynting vector $\vec{S} = \vec{E}\times\vec{B}/\mu_0$.

Frequently Asked Questions
1. How is potential difference across inductor calculated? ⌄
$V_L = L(di/dt)$. Inductor voltage depends on RATE of change of current, not current itself. Here $L = 2$ H, $di/dt = 1$ A/s, so $V_L = 2\times1 = 2$ V. If current were steady ($di/dt = 0$): $V_L = 0$ (inductor is just a wire in DC steady state). During transient: inductor acts as EMF source opposing current change.
2. What if current were decreasing instead of increasing? ⌄
If $di/dt = -1$ A/s (current decreasing): $V_L = L(di/dt) = 2(-1) = -2$ V. The inductor now acts as a source (releases stored energy), helping to maintain current. $V_A - V_B = iR + L(di/dt) = 10 + (-2) = 8$ V. The potential difference is less when current is decreasing.
3. What is the energy stored in the inductor at this instant? ⌄
$U = \frac{1}{2}Li^2 = \frac{1}{2}(2)(2)^2 = 4$ J. The energy is stored in the magnetic field of the inductor. As current increases from 2 A: more energy is stored. Rate of energy storage = $Li(di/dt) = 2\times2\times1 = 4$ W at this instant.
4. What is the time constant of this RL circuit? ⌄
$\tau = L/R = 2/5 = 0.4$ s. This means: current rises to 63.2% of final value in 0.4 s after switch closed. Effectively reaches final value in $\approx 5\tau = 2$ s. At $t = \tau = 0.4$ s: $i = 0.632(\mathcal{E}/R)$. Smaller $L$ or larger $R$ → faster response (smaller $\tau$). This is the electrical equivalent of the mechanical time constant in damped oscillations.
5. What is Lenzs law and how does it apply here? ⌄
Lenzs law: induced EMF (and resulting current) always opposes the change that caused it. Here: current is increasing ($di/dt > 0$). Inductor creates back-EMF opposing this increase. The back-EMF $= L(di/dt) = 2$ V acts in opposition to the driving voltage. This is why $V_A - V_B = 12$ V (you need 12 V to drive 2 A through 5 Ω resistance AND overcome the 2 V back-EMF of the inductor).
Previous Questions
Q.
Oxygen cylinder volume 30L gauge pressure drops mass of oxygen withdrawn
Physics · Answer: 141 g
Q.
Bus service speed interval scooty 60kmh same opposite direction observes
Physics · Answer: T=15 min, v=120 km/h
Q.
Ball dropped 40m rises 10m impulse exerted by floor on the ball
Physics · Answer: 21 N·s
Q.
P=a3b2 sqrt c d errors 1% 3% 2% 4% percentage error in P 13%
Physics · Answer: 13%
Q.
Sun expands twice radius no external torque new time period 108 days
Physics · Answer: 108 days