Current Electricity
Current Electricity for JEE Main & Advanced
Ohm's Law and Circuit Analysis
Current, Drift Velocity, Ohm's LawTopic 1
Electric Current: Rate of flow of charge. $I = dq/dt$. SI unit: ampere (A) = C/s. Scalar (though has direction of conventional positive charge flow).
Current Density $\vec{J} = I/A$ (A/m²); a vector.
Drift Velocity: Average velocity acquired by electrons under electric field $E$: $$v_d = \frac{eE\tau}{m}$$ where $\tau$ = relaxation time (mean time between collisions).
Current in terms of drift velocity: $$I = neAv_d$$ where $n$ = number density of free electrons, $A$ = cross-section.
Mobility: $\mu = v_d/E = e\tau/m$, units m²/(V·s).
Ohm's Law: $V = IR$ where $R$ = resistance (ohm, Ω). Holds for ohmic conductors at constant temperature.
Resistivity: $\rho = RA/L$. Units: Ω·m. Conductivity $\sigma = 1/\rho$.
$$R = \rho L/A, \quad \rho = \frac{m}{ne^2\tau}$$
Temperature dependence: $R_T = R_0(1 + \alpha\Delta T)$, where $\alpha$ = temperature coefficient. For metals $\alpha > 0$ (resistance increases with $T$); for semiconductors $\alpha < 0$ (decreases).
Color code for resistors (4-band): first two bands = significant digits, third = multiplier, fourth = tolerance. Mnemonic: BB ROY GBV GW.
A copper wire has $n = 8.5 \times 10^{28}$ electrons/m³, cross-section $1$ mm², current $1$ A. Find drift velocity.
Show solution
$$v_d = \frac{I}{neA} = \frac{1}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}$$ $$= \frac{1}{1.36 \times 10^4} \approx 7.35 \times 10^{-5} \text{ m/s}$$
Final Answer: $v_d \approx 7.35 \times 10^{-5}$ m/s (very slow!).
A wire of resistance $4$ Ω is stretched to twice its length. Find new resistance.
Show solution
Volume constant: $AL = A'L'$. With $L' = 2L$: $A' = A/2$. $$R' = \rho L'/A' = \rho(2L)/(A/2) = 4\rho L/A = 4R = 16 \text{ Ω}$$
Final Answer: $R' = 16$ Ω.
SI unit of current density:
Ohm's law gives:
The resistance of a metal wire with temperature:
Drift velocity of electrons in a wire is of the order:
The resistivity of a conductor depends on:
Combinations of Resistors and Kirchhoff's LawsTopic 2
Combinations of Resistors:
| Type | Equivalent | Current/Voltage |
|---|---|---|
| Series | $R_{\text{eq}} = R_1 + R_2 + \ldots$ | Same current; voltage divides as $V_i = IR_i$ |
| Parallel | $\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$ | Same voltage; current divides as $I_i = V/R_i$ |
| Two resistors in parallel | $R_{\text{eq}} = R_1R_2/(R_1+R_2)$ | — |
| $n$ equal resistors in parallel | $R/n$ | — |
Power and Energy:
| Quantity | Formula |
|---|---|
| Power dissipated | $P = VI = I^2R = V^2/R$ |
| Energy consumed | $W = Pt$ |
In series: $P \propto R$; in parallel: $P \propto 1/R$.
For combination of two bulbs: in series, lower wattage bulb glows brighter; in parallel, higher wattage glows brighter.
Kirchhoff's Laws:
- Junction (Current) Rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$ at any junction. (Conservation of charge.)
- Loop (Voltage) Rule: Sum of EMFs and IR drops around any closed loop = 0. (Conservation of energy.)
Sign Convention for Loop Rule:
- Going through battery from $-$ to $+$: $+\varepsilon$ (gain); reverse: $-\varepsilon$
- Going through resistor in direction of current: $-IR$ (drop); opposite: $+IR$
Wheatstone Bridge (balanced): $P/Q = R/S$, no current through galvanometer.
Three resistors of $2$ Ω, $4$ Ω, $6$ Ω are connected in parallel. Find equivalent resistance.
Show solution
$$\frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6+3+2}{12} = \frac{11}{12}$$ $$R_{\text{eq}} = 12/11 \approx 1.09 \text{ Ω}$$
Final Answer: $R_{\text{eq}} = 12/11$ Ω $\approx 1.09$ Ω.
Two bulbs of $40$ W and $100$ W (both rated $220$ V) connected in series across $220$ V supply. Which is brighter? Power consumed?
Show solution
Resistance: $R = V^2/P$. $R_{40} = 220^2/40 = 1210$ Ω, $R_{100} = 220^2/100 = 484$ Ω. In series, same current; bulb with higher $R$ dissipates more ($P = I^2R$). So $40$ W bulb is brighter. Current $I = 220/(1210 + 484) = 220/1694 \approx 0.13$ A. $P_{40} = I^2R_{40} = 0.13^2 \times 1210 \approx 20.5$ W; $P_{100} = 0.13^2 \times 484 \approx 8.2$ W.
Final Answer: $40$ W bulb glows brighter; $P_{40} \approx 20.5$ W.
Two equal resistors $R$ each are in parallel. Equivalent resistance:
Kirchhoff's junction rule is based on:
A heater coil of $4$ Ω draws $5$ A. Power dissipated:
In a balanced Wheatstone bridge with arms $P, Q, R, S$:
Two bulbs of $25$ W and $100$ W (same voltage rating) are connected in parallel across the rated voltage. The one which glows brighter:
Measuring Instruments and Special Circuits
Wheatstone Bridge, Potentiometer, Meter BridgeTopic 1
Wheatstone Bridge: Four resistors $P, Q, R, S$ in a quadrilateral, with battery on one diagonal and galvanometer on the other. Balance condition: No current through galvanometer when: $$\frac{P}{Q} = \frac{R}{S}$$
Used to measure unknown resistance precisely.
Meter Bridge: Practical realization of Wheatstone bridge with a $1$-m wire of uniform resistance. With unknown $X$ in one gap and standard $R$ in the other; balance length $\ell$ from $X$ side: $$X = R \cdot \frac{\ell}{100 - \ell}$$
Potentiometer: Long wire of uniform cross-section with sliding jockey. Measures EMF or potential difference without drawing current.
- Potential gradient: $k = V/L$ V/m (where $V$ = voltage applied, $L$ = length of wire)
- Balance condition: $E_{\text{unknown}} = k\ell$, where $\ell$ = balance length
- Comparison of EMFs: $E_1/E_2 = \ell_1/\ell_2$
- Measurement of internal resistance: $r = R[(l_1 - l_2)/l_2]$, where $l_1$ = balance length with open circuit, $l_2$ = balance length with $R$ in shunt.
Sensitivity of potentiometer increases with: smaller potential gradient (longer wire, smaller current).
Galvanometer: Detects small currents. Modified to ammeter (low resistance shunt in parallel) or voltmeter (high resistance multiplier in series).
In a meter bridge, an unknown resistance is in the left gap and $5$ Ω in the right gap. Balance length from left = $40$ cm. Find unknown.
Show solution
$$X = R \cdot \frac{\ell}{100-\ell} = 5 \cdot \frac{40}{60} = \frac{200}{60} = 10/3 \approx 3.33 \text{ Ω}$$
Final Answer: $X = 10/3$ Ω $\approx 3.33$ Ω.
A potentiometer wire is $10$ m long with potential gradient $0.1$ V/m. Find the EMF balanced at length $300$ cm.
Show solution
$$E = k\ell = 0.1 \times 3 = 0.3 \text{ V}$$
Final Answer: $E = 0.3$ V.
In a Wheatstone bridge, the condition for null deflection:
The potentiometer measures EMF accurately because:
To convert galvanometer to ammeter, connect:
Sensitivity of potentiometer can be increased by:
Meter bridge is based on principle of:
Cells, EMF, Internal Resistance, PowerTopic 2
EMF ($\varepsilon$): Total energy supplied by source per unit charge (open circuit voltage). Units: volt.
Internal Resistance ($r$): Resistance offered by the cell itself. $$V_{\text{terminal}} = \varepsilon - Ir$$ On open circuit ($I = 0$): $V = \varepsilon$. On short circuit: $I_{\max} = \varepsilon/r$.
Cells in Series ($n$ cells, each $\varepsilon, r$, external $R$): $$I = \frac{n\varepsilon}{R + nr}$$
Cells in Parallel ($m$ cells in parallel): $$I = \frac{\varepsilon}{R + r/m}$$
Mixed grouping ($n$ in series, $m$ rows in parallel): maximum current for $R = nr/m$.
Power Dissipation: $P = I^2R$. Maximum power transferred to external resistance when $R = r$ (impedance matching), with $P_{\max} = \varepsilon^2/(4r)$.
Power in Heating Effect (Joule's law): $H = I^2Rt$. Practical unit: kWh ($1$ kWh = $3.6 \times 10^6$ J).
Charging: When cell is charged, terminal voltage = $\varepsilon + Ir$.
Discharge time: For battery of capacity $Q_{\text{Ah}}$ at current $I$, time $t = Q/I$ hours.
A battery of EMF $12$ V and internal resistance $0.5$ Ω drives a current $2$ A through external $R$. Find $R$ and terminal voltage.
Show solution
$$\varepsilon = I(R+r) \implies R = \varepsilon/I - r = 12/2 - 0.5 = 5.5 \text{ Ω}$$ $$V = \varepsilon - Ir = 12 - 2(0.5) = 11 \text{ V}$$
Final Answer: $R = 5.5$ Ω; $V = 11$ V.
Find condition for maximum power delivered by battery ($\varepsilon, r$) to external $R$. Calculate that power.
Show solution
$$P = I^2R = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2}$$ $dP/dR = 0$ gives $R = r$. $$P_{\max} = \frac{\varepsilon^2 r}{(2r)^2} = \frac{\varepsilon^2}{4r}$$
Final Answer: $R = r$; $P_{\max} = \varepsilon^2/(4r)$.
The EMF of a cell:
Maximum power is transferred to external $R$ when:
Three identical cells of EMF $\varepsilon$ and internal resistance $r$ in parallel; equivalent EMF and resistance:
The energy consumed by a $100$ W bulb in $10$ hours:
The terminal voltage of a cell when current $I$ flows through external $R$:
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