NEET Physics Questions: Current Electricity
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Find the value of r if I = 1 A and potential difference across PQ is 1V
Vsq=i3×1+rI3
where Vsq=26/
Therefore, r = 25Ω
Two wires of equal lengths, equal diameters and having resistivities $ \rho _1 and \rho_2 $ are connected in series .The equivalent resistivity of the combination is....
$R = R_1 + R_2 = { \rho_1 L \over A} +{ \rho _2 L \over A } = { L \over A} (\rho_1 + \rho_2) .......(1)$
$ R = { \rho (2L) \over A} ......(2)$
$ Equation (1) and (2) gives = \rho = 1/2 (\rho_1 + \rho_2) $
$ I_1 = {12 V \over 4 \Omega } = 3A$
The effective resistance of a n number of resistors connected in parallel in x ohm. When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is....
$ {1 \over R_1 }+ {1 \over R_2 }+.....+{1 \over R_{n-1} }+{1 \over R_n } = {1 \over x } .....(1)$
$ \Rightarrow If the n^{th} resistor is removed , then $
$ {1 \over R_1 }+ {1 \over R_2 }+.....+{1 \over R_{n-1} }+{1 \over R_n } = {1 \over y } .....(2)$
Subtracting (2) form (1), we have
$ {1 \over R_n }= {1 \over x } - { 1 \over y}$
$$Which given {R_n} = {xy \over (y-x)} which is choice (B)$$
when a cell is connected to a resistance $R_1$ the rate at which heat is generated in it is the same as when the cell is connected to a resistance $R_ 2 < R_1$ the internal resistance of the cell is....
$ I = { E \over (R_1 + r )} \Rightarrow Q_1 = I^2 R_1 = [ { E \over (R_1 + r)}]^2 R_1$
$ Q_2 = [ { E \over (R_2 + r)}]^2 R_2$
$Equating Q_1 and Q_2 simplifying, we get \,r = \sqrt { R_1 R_2} $
Two batteries each of emf 2V and internal resistance $1 \Omega $ are connected in series to a resistor R. Maximum Possible power consumed by the resistor = ....
$ Max power = n ({ E^2 \over 4r})$
In an experiment to measure the internal resistance of a cell by a potentiometer it is found that all the balance points at a length of 2m when the cell is shunted by a 5 ohm resistance and is at a length of 3m when the cell is shunted by a 10 ohm resistance, the internal resistance of the cell is then :
$ r = ({ l_1 \over l_2} -1) R$
Two wires of the metal have the same length but their cross-sections are in the ratio 3:1 They are joined in series: The resistance of the thicker wire is $ 10 \Omega $ . The total resistance of the combination will be
$ R_1 = { \delta l \over 3 } = 10 \Omega ,
R _2 ={ \delta l \over 1} = 30 \Omega $
$ Rs = R_1 + R_2 = 10 \Omega + 30 \Omega = 40 \Omega $
The ratio of angular speeds of minute hand and hour hand of a watch is
ω=2π/T
A wire 50cm long and $1 mm^2$ in cross-section carries a curent of 4A when connected to a 2V battery. The resistivity of the wire is:
A parallel combination of three resistors takes a current of 7.5 A form a 30 V supply, It the two resistors are $ 10 \Omega $ and $ 12 \Omega $. find which is the third one?
$ I = 7.5 A , V = 30 V, R_1 = 10 \Omega , R_2 = 12 \Omega , R_3 = ? $
$ R_{net} = { V \over I}$
$ \Rightarrow { 1 \over Rnet }= {I \over V} = { 1 \over R_1 } + { 1 \over R_2 } + { 1 \over R_3 } $
A potentiometer wire has length 10 mand resistance $ 20 \Omega $. A 2.5V battery of negligible internal resistance is connected across the wire with an $ 80 \Omega $ series resistance. The potential gradient on the
$ V = ( {E \over R_S + R_i}) Rl$
$ =({ 2.5 \over 80+20}) \times 20= 0.5 volt $
$ Potential gradient = { V/ l} = { 0.5 volt / 10 metre } = 5 \times 10 ^ {-2} Vm^{-1} = 5 \times 10^{-5} V /mm$
The drift velocity of free electrons through a conducting wire of radius r, carrying current I, is if the same current is passed through a conductor of radius 2r what will be the drift velocity?
$ I = nAV_dq$
$V_d \alpha 1/ r^2$
$ \therefore V_d^1 = Vd /4 $
A carbon resistor has a set of coaxial coloured rings in the order brown, violet brown and silver. The value of resistance (in ohms) is.
A cross a wire of length l and thickness d, a p.d of V is applied. If the p.d is doubled the dirft velocity becomes....
$ I =V/R= nAV_dq$
$ \therefore V_d= V $
The masses of three wires of copper are in the ratio of 1:3:5 and their lengths are in the ratio of 5:3:1. The ratio of their electrical resistance is:
$R \alpha {l^2 \over m} $
$\therefore R_1 : R_2:R_3= {25 \over 1} : { 9 \over 3} : { 1 \over 5 } =125 : 15:1$