Three cells are connected in parallel with their like poles connected together with wires of negligible resistance. If the emfs of the cells are 2, 1 an 4 V respectively and their internal resistances are 4,3 and `2Omega` respectively, find the current through each cell.
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Two resistance connected in parallel across a cell of negligible internal resistance, use 4.5 times the power that they would use, when in series acro
`P_(p)=(V^(2)(R_(1)+R_(2)))/(R_(1)R_(2))` `P_(s)=(V^(2))/(R_(1)+R_(2))` `P_(P)=4.5Ps` `(R_(1)+R_(2))/(R_(1)R_(2))=(9)/(2(R_(1)+R_(2))) because (R_(1)=8.2)` `(8+R_(2))/(8R_(2))=36 R_(2)` `R_(2)^(2)-20R_(2)+ 6 4=0` `R_(2)=16,4`
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A 10 wire potentiometer has first five wires of cross sectional radius `r` and next five wires of radius `2r`. The wires are uniform and made of same
Correct Answer - (a) 3.125 (b) 4.688 m (c) 7.5 m
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Two batteries of emfs 24V and 6V and internal resistance `2Omega` and `1Omega` respectively are joined in parallel, negative to negarive by a resistor
Correct Answer - 20 V, 8V
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The sides of a square ABCD are formed of wires, each of resistance r and B is joined to D by a wire of resistance 2r. Find the equivalent resistance o
Correct Answer - i) 2/3r, ii) r, iii) 2/3r
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A solenoid having a resistance of `5Omega` and self-inductance of 4H is connected to a battery of emf 10V and negligible resistance. After how long wi
Correct Answer - 0.555s
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An accumulator of emf `2 V` and negligible internal resistance is connected across a uniform wire of length 10 m and resistance `30 Omega` The appropr
Correct Answer - 7.5 m,8.75 m,6.25 m
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A 10 wire potentiometer has a resistance of 10` Omega` and is connected to fan accumulator of 2 V and negligible internal resistance.There are two res
Correct Answer - Put `R_(1)` =1018 `Omega`(1000 times emf of Cd-cell),adust `R_(2)` till no deflection ocurs in the galvanometer , `R_(2)`=972 `Omega`;5.1 m
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In the cirucit (figure6.3) resistance `R_(1) and R_(2)` are known and cells ar of emf `e_(1) and e_(2)` and negligible internal resistance. At what va
Correct Answer - `R=R_(1)//R_(2//)(R_(1)+R_(2));p_(max)=(epsilon_(1)R_(2)+epsilon_(2)R_(1))/(4R_(1)+R_(2)(R_(1)+R_(2)))`
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The resistance of two conductors in series is `18 Omega` and the resistance becomes `4 Omega` when connected in parallel. Find the resistance of indiv
Correct Answer - `12 Omega, 6 Omega`
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Three identical cells each of emf 2 V and unknown internal resistance are connected in parallel. This combination is connected to a `5 Omega` resistor
Correct Answer - `5 Omega`
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