Practice free Current Electricity (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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} $ $\therefore R_1 : R_2:R_3= {25 \over 1} : { 9 \over 3} : { 1 \over 5 } =125 : 15:1$
Resistivity of material of a conducting wire is $ 4 \times 10^{–8} \Omega m$ volume of the wire is $4m^3$ and its resistance is$ 4 \Omega $ Therefore its length will be.
$ l^2 ={ RV \over \rho}$
How would you arrange 48 cells each of e.m.f 2V and inteanal resistance $1.5 \Omega $ so as to pass maximum current through the external resistance of $2 \Omega$ ?
[{No of rows}/{no of cells in each row}] = (m/n) = (r/R) where r = internal resistance of each cell
R = external resistance
∴ (m/n) = {1.5/2} = (3/4)
also total number of cells = 48 m × n = 48
∴ (3/4) n × n = 48 n2 = {(48 × 4)/3} n = 8
∴ m = (48/8) = 6
∴ 6 rows with 8 cell in each of row
How many dry cells, each of emf 1.5V and internal resistance $0.5 \Omega $, much be joined in series with a resistor of $ 20 \Omega $ to give a current of 0.6A in the circuit ?
$ l = { n \varepsilon \over R +nr } \Leftrightarrow 0.6 = { n \times 1.5 \over 20 \times 0.5 n } \Leftrightarrow 12 + 0.3 n = 1.5 n \therefore 12 = 1.2 n $ $ \therefore n = 10 $cells to be connected in series
Two electric bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a constant voltage source the power dissipated in them have the ratio.
$ {P_1/P_2 } = {V^2 / R_1 \over V^2 / R_2 } = {R_2 \over R_1} = 2 /1 $
An electric kettle has two coils. when onc of them is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes If the two coils are connected in series the time taken to boil water in the kettle is:
$H_1 = { V^2 \over R_1 } t_1$ $ R_1 = { V^2 \over H_1} t_1 and R_2 = { V^2 \over H_2 } t_2$ $ { V^2 \over H_1 } = { V^2 \over H_2 } = const \therefore R\, \alpha \,t $ $ Now R_s = R_1 + R_2 $ t = 6 + 3 = 9 Minutes
An electric kettle has two coils when one of these is switched on. the water in the kettle boils in 6 minutes. When the other coil is switched on, boils in 3 minutes If the two coils are connected in parallel, the time taken to boil water in the kettle is
$R_p = { R_1 R_2 \over R_1 + R_2 } $ $ t = {6 \times 3 \over 6 +3} = 2 minutes $
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