Vardaan Learning Institute

Ans: (b) Volt
S.I unit is Volt (V).

Ans: (b) $IR^2$
Formulas for power are $P=VI, P=I^2R, P=V^2/R$. $IR^2$ is incorrect.

Ans: (a) 6 $\Omega$
$R_eq = R_1 + R_2 = 2 + 4 = 6 \Omega$.

Ans: (c) Kilowatt-hour (kWh)
1 unit = 1 kWh.

Ans: (c) V = IR
Ohm's law states $V \propto I$ or $V=IR$.

Ans: (a) Ammeter
Ammeter measures electric current.

Ans: (b) Tungsten
High melting point ($3380^\circ C$).

Ans: (b) Fuse
Breaks circuit on overloading.

Ans: (b) Potential Difference
Voltage remains same across parallel components.

Ans: (c) Electric current
$I = Q/t$.

Ans: (a) Both A and R are true and R is the correct explanation of A.
Tungsten has high melting point ($3380^\circ C$), so it doesn't melt at high temperatures when bulb glows.

Ans: (a) Both A and R are true and R is the correct explanation of A.
High resistivity and oxidation resistance make them ideal for heating elements.

Ans: (a) Both A and R are true and R is the correct explanation of A.
Same current flows through all components in a single path series circuit.

Ans: (d) A is false but R is true.
Fuse is always connected in SERIES, not parallel. The reason (R) correctly describes fuse function.

Resistance depends on:
1. Length of conductor ($R \propto L$)
2. Area of cross-section ($R \propto 1/A$)
3. Nature of material (Resistivity $\rho$)
4. Temperature

$R = 2 \times 10^{-2} \Omega$, $L = 2$ m, $A = 1.7 \times 10^{-6} m^2$.
$R = \rho \frac{L}{A} \Rightarrow \rho = \frac{RA}{L}$.
$\rho = \frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2} = 1.7 \times 10^{-8} \Omega m$.

1. If one appliance fails (e.g., bulb fuses), the circuit is broken and all other appliances stop working.
2. Total resistance increases, reducing current.
3. Voltage is divided across appliances, so they don't get 220V.

Rated: P=100W, V=220V.
$R = \frac{V^2}{P} = \frac{220 \times 220}{100} = 484 \Omega$ (Resistance remains constant).
New V = 110V.
New Power $P' = \frac{V'^2}{R} = \frac{110 \times 110}{484} = 25$ W.

$Q = ne \Rightarrow n = Q/e$.
$Q = 1$ C, $e = 1.6 \times 10^{-19}$ C.
$n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons.

Power P = 400 W = 0.4 kW.
Time t = 8 hrs/day $\times$ 30 days = 240 hours.
Energy E = $P \times t = 0.4 \times 240 = 96$ kWh.
Cost = $96 \times 3.00 = 288$ Rs.

(i) In parallel, V is same, $I = I_1 + I_2 + I_3$.
$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$.
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.

(ii) $R_1=5, R_2=10, R_3=30$, V=12V.
(a) Current:
$I_1 = V/R_1 = 12/5 = 2.4$ A.
$I_2 = V/R_2 = 12/10 = 1.2$ A.
$I_3 = V/R_3 = 12/30 = 0.4$ A.
(b) Total Current $I = 2.4 + 1.2 + 0.4 = 4.0$ A.
(c) Total Resistance $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{6+3+1}{30} = \frac{10}{30} = \frac{1}{3} \Rightarrow R_p = 3 \Omega$.

(i) Heat produced in a resistor is directly proportional to square of current ($I^2$), resistance ($R$), and time ($t$). $H = I^2Rt$.
(ii) $R=20 \Omega, I=5 A, t=30 s$.
$H = 5^2 \times 20 \times 30 = 25 \times 600 = 15000$ J or 15 kJ.
(iii) The heating element has very high resistance, so large heat ($I^2Rt$) is produced making it glow. The connecting cord acts as a conductor with negligible resistance, so heat produced is negligible.

(i) Silver (lower resistivity $1.60 \times 10^{-8}$) is a better conductor than Copper ($1.62 \times 10^{-8}$).
(ii) Nichrome (High resistivity, doesn't oxidise easily).
(iii) Alloys have higher resistivity than constituent metals and do not oxidise (burn) readily at high temperatures.

(i) To protect circuits and appliances from high current (overloading/short-circuiting).
(ii) Series.
(iii) Tin-lead alloy (solder) has a low melting point, so it melts easily when current exceeds safe limit.