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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 $
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$
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