Potential & Potential Difference: Flow of charge is driven by potential difference (like water level difference). Positive charge moves from higher to lower potential; electrons move opposite.

Current: \( I=\dfrac{Q}{t} \). Unit A (ampere); \(1\ \text{A} = 1\ \text{C s}^{-1}\). Charge of electron \(=1.6\times10^{-19}\ \text{C}\).

Ohm’s Law: If physical state is constant, \(V \propto I \Rightarrow V=IR\). Here \(R\) is resistance (Ω).

Resistance & Resistivity: For a wire \(R=\rho\dfrac{L}{A}\), where \(L\) = length, \(A\) = cross-sectional area, \(\rho\) (ohm–metre) depends only on material (at fixed temperature).

Series Connection: Same current, \(R_s=R_1+R_2+\cdots\); p.d. divides.

Parallel Connection: Same p.d., \(\displaystyle \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots\); currents add.

Instruments: Ammeter in series (low \(R\)); Voltmeter in parallel (high \(R\)).

Domestic Wiring: Live (brown/red), Neutral (blue/black) ~ \(220\ \text{V}\) in India; Earth (green/yellow) for safety. Appliances are connected in parallel.

Safety: Fuse/MCB breaks circuit on overcurrent; dry hands; insulated footwear; never touch a shocked person—switch off supply first.

1) SI unit of electric charge?

Coulomb (C).

2) SI unit of current?

Ampere (A), \(1\,\text{A}=1\,\text{C s}^{-1}\).

3) Symbol and unit of resistance?

\(R\), unit Ohm \((\Omega)\).

4) Mathematical form of Ohm’s law?

\(\boxed{V=IR}\).

5) Define 1 volt.

\(1\ \text{V}=1\ \text{J C}^{-1}\) potential difference.

6) Define 1 ampere.

Current when \(1\ \text{C}\) passes per second.

7) Conventional current direction?

From + to − terminal (opposite to electron flow).

8) Electron charge (magnitude)?

\(|e|=1.6\times10^{-19}\ \text{C}\).

9) Resistivity unit?

\(\Omega\ \text{m}\) (ohm–metre).

10) Effective resistance in series?

Sum of individual resistances.

11) Effective resistance in parallel?

Reciprocals add: \(\frac1{R_p}=\sum \frac1{R_i}\).

12) Meter to measure p.d.?

Voltmeter (in parallel).

13) Meter to measure current?

Ammeter (in series).

14) Household potential difference in India?

\(\approx 220\ \text{V}\) between live and neutral.

15) Which connection for equal p.d. across devices?

Parallel.

16) Which connection increases total resistance?

Series.

17) Formula for current using charge and time?

\(I=\dfrac{Q}{t}\).

18) Area of a wire of diameter \(d\)?

\(A=\pi\left(\dfrac{d}{2}\right)^2\).

19) What is a fuse?

Safety device that melts on excessive current.

20) Colour of earth wire?

Green/Yellow-green.

1) Define electric potential difference between A and B.

Work done to move unit positive charge from A to B: \(V_{AB}=\dfrac{W}{Q}\).

2) Why do electrons move from − to + in a circuit?

They are negatively charged and are pushed by the electric field towards higher potential.

3) State Ohm’s law.

At constant physical conditions, \(V\propto I\Rightarrow V=IR\).

4) On what factors does resistance of a wire depend?

Material (ρ), length \(L\) (∝), area \(A\) (∝ \(1/A\)), and temperature.

5) Write relation connecting \(R, \rho, L, A\).

\(R=\rho\dfrac{L}{A}\).

6) Why is voltmeter of high resistance?

So that negligible current flows through it; does not disturb circuit.

7) Why is ammeter of low resistance?

To allow full circuit current with minimal drop; connected in series.

8) State two differences between series and parallel connections.

Series: same current, \(R\) increases. Parallel: same p.d., \(R\) decreases.

9) What is conventional current?

Assumed flow of positive charge from + to − terminal.

10) Why are domestic appliances connected in parallel?

Each gets same rated p.d.; independent control; failure of one doesn’t stop others.

11) Define 1 ohm.

Resistance when \(1\ \text{A}\) flows for \(1\ \text{V}\) across the conductor.

12) What is a superconductor?

Material with nearly zero resistance below critical temperature.

13) If \(I=0\), what is the current through each series resistor?

Zero—the same current flows through all series components.

14) State unit conversions: mA and μA.

\(1\ \text{mA}=10^{-3}\ \text{A},\ 1\ \mu\text{A}=10^{-6}\ \text{A}\).

15) Give the formula of effective resistance of two resistors in parallel.

\(\displaystyle R_p=\frac{R_1R_2}{R_1+R_2}\).

16) Which wire colour is “live” in India?

Brown/Red.

17) Define current density briefly (idea only).

Current per unit area normal to flow: \(J=\dfrac{I}{A}\).

18) What happens to \(R\) if diameter doubles?

Area becomes 4×, so \(R\) becomes \(\dfrac{1}{4}\) (for same \(L,\rho\)).

19) Which has lower effective resistance: series or parallel?

Parallel—always less than the least individual \(R\).

20) What is the safe step when someone gets an electric shock?

Switch off main supply first / use dry wooden stick to separate; do not touch directly.

1) Explain the waterfall analogy for potential difference.

Water flows from high to low level; similarly charge flows due to p.d. High–low potential acts like level difference.

2) Define current and derive its unit.

\(I=\dfrac{Q}{t}\); unit \(\text{C s}^{-1}=\text{A}\). 1 A flows when 1 C passes a cross-section each second.

3) State conditions for validity of Ohm’s law.

Physical state must remain constant—temperature, length, area, and material unchanged.

4) Why does resistance cause heating?

Collisions of electrons with ions transfer energy to lattice → Joule heating.

5) How does wire length and area affect \(R\)?

\(R\propto L\) (longer → more collisions), \(R\propto 1/A\) (thicker → more lanes, less opposition).

6) Give two uses of high-resistance voltmeter and low-resistance ammeter.

Voltmeter: measures p.d. without drawing current; Ammeter: measures full circuit current with negligible drop.

7) Why are bulbs dimmer in series?

Series increases \(R\) → current reduces → power \(P=VI=I^2R\) per bulb drops.

8) Show that \(1\ \Omega = 1\ \text{V A}^{-1}\).

From \(V=IR\Rightarrow R=\dfrac{V}{I}\); units give \(\Omega=\text{V}/\text{A}\).

9) Why are domestic loads kept parallel?

Each gets rated voltage, independent switching, constant brightness; fault in one doesn’t affect others.

10) Write the formula for resistivity and its unit from \(R=\rho L/A\).

\(\rho=RA/L\); units \(\Omega\ \text{m}\).

11) Two resistors \(R_1, R_2\) in series: current and p.d. relation?

Current same \(I\). p.d. divides: \(V_1=IR_1,\ V_2=IR_2\).

12) Two resistors \(R_1, R_2\) in parallel: current and p.d. relation?

p.d. same \(V\). Currents \(I_1=\dfrac{V}{R_1}, I_2=\dfrac{V}{R_2}\); \(I=I_1+I_2\).

13) What is the purpose of earthing?

Provides a low-resistance path to ground for fault current; prevents metal body from becoming live.

14) Define milliampere and kilovolt.

\(1\ \text{mA}=10^{-3}\ \text{A}\); \(1\ \text{kV}=10^3\ \text{V}\).

15) Why is copper a better conductor than nichrome?

Lower resistivity (\(\rho_{\text{Cu}}\approx1.7\times10^{-8}\ \Omega\text{m}\) vs nichrome \(\sim10^{-6}\ \Omega\text{m}\)).

16) How does temperature affect metal resistance?

Generally increases (more lattice vibrations → more collisions).

17) State two precautions while using electricity at home.

Switch off before cleaning; dry hands & insulated footwear; keep sockets away from children.

18) Explain why voltmeter is connected across the resistor.

It must measure p.d. between two points of the resistor; connected in parallel.

19) Give the relation between power and current, voltage.

\(P=VI=I^2R=\dfrac{V^2}{R}\).

20) What is the advantage if one bulb fuses in a parallel circuit?

Other bulbs keep glowing because alternate paths remain intact.

A. By which method are the appliances connected?

Parallel connection.

B. What must be the potential difference across individual appliances?

Same as supply p.d. (≈ \(220\ \text{V}\) in India) across each appliance.

C. Will the current passing through each appliance be the same? Justify.

No; current depends on each device’s resistance \(I=\dfrac{V}{R}\) though p.d. is same.

D. Why are the domestic appliances connected this way?

Parallel ensures rated voltage for each, independent control, and equal brightness/performance.

E. If the TV stops working, will others also stop? Explain.

No. In parallel, other branches carry current independently; the circuit remains complete.

Which law can be proved using ammeter in series and voltmeter across the resistor?

Ohm’s Law: Plot \(V\) vs \(I\) keeping conductor’s physical state constant; a straight line through origin verifies \(V\propto I\).

A. Which method should he use to connect the bulbs?

Series connection (to increase effective resistance, hence limit current).

B. Characteristics for this way of connecting?

Same current flows through both; effective resistance \(R_s=R_1+R_2\) higher → safer current; brightness reduces.

Given \(V\)–\(I\) pairs: (4, 0.5), (5, 0.625), (6, 0.75), (9, 1.125), (13.5, 1.6875), (11.25, 1.40625). Find average resistance, nature of graph, and name the law.

For each pair \(R=\dfrac{V}{I}\) gives \(8\ \Omega\) in all cases ⇒ average \(R=8\ \Omega\). Graph of \(V\) vs \(I\) is a straight line through origin ⇒ verifies Ohm’s Law.

The resistance of a conductor of length \(x\) is \(r\). Area of cross-section is \(a\). What is its resistivity and unit?

\(\displaystyle \rho=\dfrac{RA}{L}=\dfrac{r\,a}{x}\). Unit: \(\Omega\ \text{m}\).

Discuss current in cases: (a) both S1 and S2 closed; (b) both open; (c) S1 closed, S2 open.

(a) Both closed: All intended paths complete; current divides according to branch resistances (parallel parts active).
(b) Both open: Open switches break paths; no current in open branches; only paths not interrupted (if any) will carry current, else zero.
(c) S1 closed, S2 open: Only the branch controlled by S1 conducts; branch under S2 is open → no current there.

a) Same current \(I\) through \(x_1,x_2,x_3\)

Series.

b) \(x\) (effective) larger than each of \(x_1,x_2,x_3\)

Series.

c) \(x\) smaller than each of \(x_1,x_2,x_3\)

Parallel.

d) Same potential difference across \(x_1,x_2,x_3\)

Parallel.

e) \(x=x_1+x_2+x_3\)

Series.

f) \(\displaystyle \frac{1}{x}=\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\)

Parallel.

A) Nichrome wire: \(R=6\ \Omega\) at \(L=1\ \text{m}\). Find \(R\) at \(L=0.70\ \text{m}\) (same area, temperature).

\(R\propto L\Rightarrow R'=6\times\frac{0.70}{1}=4.2\ \Omega\).

B) Two resistors have \(R_s=80\ \Omega\) (series) and \(R_p=20\ \Omega\) (parallel). Find the two resistances.

Let \(R_1+R_2=80\) and \(\dfrac{R_1R_2}{R_1+R_2}=20\Rightarrow R_1R_2=1600\).
Solve: \(R_1,R_2\) are roots of \(x^2-80x+1600=0\Rightarrow (x-40)^2=0\). Hence \(R_1=R_2=40\ \Omega\).

C) Charge \(Q=420\ \text{C}\) flows in \(5\) minutes. Find current.

\(t=5\times60=300\ \text{s}\). \(I=\dfrac{Q}{t}=\dfrac{420}{300}=1.4\ \text{A}\).

1) Bulb filament \(R=1000\ \Omega\), \(V=230\ \text{V}\): find \(I\).

\(I=\dfrac{V}{R}=\dfrac{230}{1000}=0.23\ \text{A}\).

2) Wire \(L=50\ \text{cm}=0.50\ \text{m}\), radius \(r=0.5\ \text{mm}=5\times10^{-4}\ \text{m}\), \(R=30\ \Omega\). Find \(\rho\).

\(\displaystyle \rho=\dfrac{RA}{L}=\dfrac{30\cdot\pi (5\times10^{-4})^2}{0.50}=4.71\times10^{-5}\ \Omega\text{m}\) (using \(\pi\approx3.14\)).

3) Conductor with \(V=24\ \text{V}\), \(I=0.24\ \text{A}\): find \(R\).

\(R=\dfrac{V}{I}=\dfrac{24}{0.24}=100\ \Omega\).

4) Appliance with \(R=110\ \Omega\) at \(V=33\ \text{V}\): find \(I\). For same \(I\) with \(R=500\ \Omega\): find \(V\).

\(I=\dfrac{33}{110}=0.3\ \text{A}\). Then \(V=IR=0.3\times500=150\ \text{V}\).

5) Copper wire: \(\rho=1.7\times10^{-8}\ \Omega\text{m}\), \(L=1\ \text{km}=10^3\ \text{m}\), diameter \(d=0.5\ \text{mm}=5\times10^{-4}\ \text{m}\). Find \(R\).

\(A=\pi(d/2)^2=\pi(2.5\times10^{-4})^2\approx1.9635\times10^{-7}\ \text{m}^2\).
\(R=\rho\dfrac{L}{A}\approx\dfrac{1.7\times10^{-8}\times10^{3}}{1.9635\times10^{-7}}\approx86.6\ \Omega\).
(Rounded ≈ \(85\ \Omega\) if \(\pi\) or area is approximated in textbook.)

6) Series example: \(15\ \Omega,3\ \Omega,4\ \Omega\) in series—find effective \(R\).

\(R_s=15+3+4=22\ \Omega\).

7) Series example: \(16\ \Omega,14\ \Omega\) at \(V=18\ \text{V}\)—find \(I,V_1,V_2\).

\(R_s=30\ \Omega\Rightarrow I=\dfrac{18}{30}=0.6\ \text{A}\). \(V_1=0.6\times16=9.6\ \text{V}\), \(V_2=0.6\times14=8.4\ \text{V}\).

8) Parallel example: \(15,20,10\ \Omega\) in parallel—find \(R_p\).

\(\dfrac{1}{R_p}=\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{10}=\dfrac{13}{60}\Rightarrow R_p=\dfrac{60}{13}\approx4.615\ \Omega\).

9) Parallel example: \(5,10,30\ \Omega\) at \(12\ \text{V}\)—find \(I_1,I_2,I_3, I, R_p\).

\(I_1=12/5=2.4\ \text{A},\ I_2=12/10=1.2\ \text{A},\ I_3=12/30=0.4\ \text{A}\). \(I=4.0\ \text{A}\). \(R_p=V/I=12/4=3\ \Omega\).