Practice free Current Electricity (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The potential difference between the terminals of a battery is 10V and internal resistance $1 \Omega $ drops to 8V when connected across an external resistor find the resistance of the external resistor.
$ V = IR \,in I = { E \over R +r } $ $ \therefore ={ ER \over R + r } $
A heater boils 1kg of water in time $t_1$ and another heater boils the same water in time $t_2$ If both are connected in series, the combination will boil the same water in time.
At what temperature will the resistance of a copper wire be three times its value at 0°C ? (Given: temperature coefficient of resistance for copper = $4 \times 10^{-3} C^{-1}$ )
$R_\theta = R_\theta 0 [ 1 + \alpha ( \theta _2 - \theta _1 ) ] $ $ 3 = 1 + ( 4 \times 10 ^ {-3} ) T$ $ \therefore T = { 2 \over 4 \times 10 ^ {-3}} = 500 C $
The resistance of a copper coil is $4.64 \Omega $ at 40 °C and $5.6 \Omega$ at 100 °C Its resistcnce at 0° C will be
$R_{40} [1 + \alpha (\theta_2 - \theta_1)]$ $ 5.6 = 4.64[1 + a(100 -40)] =4.64+278.4 a$ $a ={5.6 -4.64 \over 278.4} = 0.0035 C^{-1}$ $ \Rightarrow _{100} = R_0(1+ \alpha T_2) = R_0 [1+(0.0035 \times 100)]$ $ 5.6 = R_0 \times 1.35$ $Ro = 4 \Omega $
There are n resistors having equal value of resistance r. First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is....
In parallal connection R minimum = r/n In Series connection R maximum= nr $ { R min \over R max } = { r \over n(n-1) } = {1 \over n^2}$
Temperature of a conductor increases by $5 ^\circ $C passing electric current for some time. The increase in its temperature when double current is passed through the same conductor for the same time is $ ^ \circ C $
$ \triangle Q\, \alpha\, I^2 $ V is same so $ P_1 = { V^2 \over R_1 } and P_2 = {V^2 \over R_2} $
Area of cross-section of two wires of same length carrying same current is in the ratio of 1 : 2. Then the ratio of heat generated per second in the wires = ....
$ I , \rho , l are equal so H \alpha {1 \over A} $ $ \therefore {H_1 \over H_2} = {A_2 \over A_1} = {2 \over 1} $
If $ \sigma_1, \sigma_2, \sigma_3$ are the conductances of three conductor then equivalent conductance when they are joined in series, will be.
$ Reff = R_1 =R_2 +R_3 \Rightarrow {1 \over \sigma_eff} = { 1 \over \sigma _2 } + { 1 \over \sigma _3} $
n resistors each of resistance r are connected to a battery of emf E and internal resistance r. Then the ratio of terminal voltage to emf of battery =....
$ V = \varepsilon-Ir , \varepsilon = V+Ir ,$ $ but V = nlr , \varepsilon = nlr +Ir = (n+1) Ir$ $\therefore {V \over \varepsilon} = { n \over n+1 } $
Masses of three conductors of same material are in the proportion of 1:2:3 their lengths are in the proportion of 3:2:1 then their resistance will be in the proportion of....
$Mass m = density \times volume = dAl$ $ \therefore A = {m \over dl} $ $ Now, R = { \rho l \over A} = { \rho l d l \over m} = { \rho d l^2 \over m } $ $ \therefore R \alpha {l^2 \over m } ( \therefore p and d = constant ) $ $ \therefore R_1 : R_2:R_3 = { 9 \over 1} : { 4 \over 2 } : { 1 \over 3 } = 27 :6:1 $
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