Current Electricity - NEET Physics Chapterwise MCQs & PYQs
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NEET Current Electricity MCQs & PYQs
Practice NEET Current Electricity Questions
Question 201:
easy
A cell can be balanced against $110\text{ cm}$ and $100\text{ cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10\text{ }\Omega$. Its internal resistance is:
(2008)
The open-circuit balancing length is $l_1 = 110\text{ cm}$ and closed-circuit balancing length is $l_2 = 100\text{ cm}$.
Using internal resistance formula $r = R \left(\frac{l_1}{l_2} - 1\right)$:
$r = 10 \cdot \left(\frac{110}{100} - 1\right) = 10 \cdot 0.1 = 1.0\text{ ohm}$.
The potentiometer is best for measuring voltage, as:
(2000)
At the balancing point, no current is drawn from the secondary circuit by the potentiometer.
Thus, it acts as an ideal voltmeter with infinite resistance.
It measures potential difference across open circuit conditions without disturbing the circuit.
A filament bulb ($500\text{ W}$, $100\text{ V}$) is to be used in a $230\text{ V}$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\text{ W}$. The value of $R$ is:
(2016 – II)
Rated current for the bulb is $I = \frac{P}{V_b} = \frac{500}{100} = 5\text{ A}$.
To operate on a $230\text{ V}$ supply, potential drop across $R$ must be $V_R = 230 - 100 = 130\text{ V}$.
Therefore, $R = \frac{V_R}{I} = \frac{130}{5} = 26\text{ }\Omega$.
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
(2017-Delhi)
A potentiometer measures the electromotive force (e.m.f.) of a cell using a null deflection method.
At the balance point, no current flows through the galvanometer, so no current is drawn from the cell being measured.
This allows it to measure the true open-circuit potential difference accurately.
A potentiometer wire is $100\text{ cm}$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\text{ cm}$ and $10\text{ cm}$ from the positive end of the wire in the two cases. The ratio of emf’s is:
(2016 – I)
When cells support each other, $E_1 + E_2 = k \cdot 50$. When they oppose each other, $E_1 - E_2 = k \cdot 10$.
Taking the ratio gives $\frac{E_1 + E_2}{E_1 - E_2} = 5$, which simplifies to $6 E_2 = 4 E_1$.
Thus, the ratio of their e.m.f.'s is $\frac{E_1}{E_2} = \frac{3}{2}$.
A potentiometer wire has length $4\text{ m}$ and resistance $8\text{ }\Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\text{ V}$, so as to get a potential gradient $1\text{ mV}$ per $\text{cm}$ on the wire is:
(2015)
Potential drop across the wire $V_w = k \cdot L = 10^{-3}\text{ V/cm} \cdot 400\text{ cm} = 0.4\text{ V}$.
Current required $I = \frac{V_w}{R_w} = \frac{0.4}{8} = 0.05\text{ A}$.
Total resistance of circuit $R_{\text{total}} = \frac{E}{I} = \frac{2}{0.05} = 40\text{ }\Omega$, so series resistance $R = 40 - 8 = 32\text{ }\Omega$.
Two cities are $150\text{ km}$ apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is $8\text{ volt}$ and the average resistance per km is $0.5\Omega$. The power loss in the wire is:
(2014)
Total resistance $R = 150 \times 0.5 = 75\Omega$. Total voltage drop $V = 150 \times 8 = 1200\text{ V}$. Power loss $P = \frac{V^2}{R} = \frac{(1200)^2}{75} = 19200\text{ W} = 19.2\text{ kW}$.
If voltage across a bulb rated $220\text{ Volt} – 100\text{ Watt}$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is:
(2012 Pre)
Power is given by $P = \frac{V^2}{R}$. For small changes, the fractional change is $\frac{\Delta P}{P} = 2\frac{\Delta V}{V}$. The percentage decrease in power will be $2 \times 2.5\% = 5\%$.
When three identical bulbs of $60\text{ watt}$, $200\text{ volt}$ rating are connected in series to a $200\text{ volt}$ supply, the power drawn by them will be:
(2004)
Resistance of one bulb is $R = \frac{V^2}{P} = \frac{200^2}{60}$. In series, total resistance $R_{\text{eq}} = 3R$. Power drawn is $P' = \frac{V^2}{3R} = \frac{1}{3} \frac{V^2}{R} = \frac{60}{3} = 20\text{ watt}$.
In India electricity is supplied for domestic use at $220\text{ V}$. It is supplied at $110\text{ V}$ in USA. If the resistance of a $60\text{ W}$ bulb for use in India is $R$, the resistance of a $60\text{ W}$ bulb for use in USA will be:
(2004)
Power $P = \frac{V^2}{R}$, which implies $R \propto V^2$ for the same power. Thus, $\frac{R_{\text{USA}}}{R_{\text{India}}} = (\frac{110}{220})^2 = \frac{1}{4}$, yielding $R_{\text{USA}} = \frac{R}{4}$.