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This is one of our many quizzes on Parallel circuits. Click on this to access more such competitive quizzes on Current Electricity.

Parallel circuits Quiz 2

physics

A circuit made of capacitors is shown in the figure below. Find the charge on each capacitor. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9c2e0ef7-1d29-4ddd-a93f-c7d8cf4e3f378981627846349369097.png"/>

 The voltage across both the arms of the parallel combination is proportional to the equivalent capacitance. 

  $\text{The equivalent capacitance for the given circuit is;} \\ {{C}_{eq}}=\dfrac{1}{\left( \dfrac{1}{C}+\dfrac{1}{C} \right)}+\dfrac{C}{1} \\ \Rightarrow {{C}_{eq}}=\dfrac{C}{2}+\dfrac{C}{1}=\dfrac{3C}{2} \\ \text{The voltage across ‘a’ is }V; \\ \text{The charge in ‘a’ is given by;} \\ \therefore {{q}_{a}}=\dfrac{CV}{1} \\ \text{The voltage across  }\!\!'\!\!\text{ b }\!\!'\!\!\text{  is given by;} \\ \Rightarrow {{V}_{b}}=\dfrac{{{C}_{c}}V}{{{C}_{b}}+{{C}_{c}}}=\dfrac{CV}{2C}=\dfrac{V}{2} \\ \text{The charge in ‘b’ is given by;} \\ \therefore {{q}_{b}}=\dfrac{CV}{2} \\ \text{The voltage across ‘c’ is given by;} \\ \Rightarrow {{V}_{c}}=\dfrac{{{C}_{b}}V}{{{C}_{b}}+{{C}_{c}}}=\dfrac{CV}{2C}=\dfrac{V}{2} \\ \therefore {{q}_{c}}=\dfrac{CV}{2}$

Calculate the energy stored in each capacitor and the total energy stored in the circuit. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6e657f12-b12e-4ff4-b8d5-64c45f7e30ec8379328669964659931.png"/>

 The potential difference across the capacitors connected in parallel is the same and the energy of the circuit will always be conserved. 

  $\text{The energy stored in a capacitor is given by;} \\ E=\dfrac{1}{2}C{{V}^{2}} \\ \text{The energy stored in each capacitor in the circuit is;} \\ \Rightarrow {{E}_{1}}=\dfrac{1}{2}\times \dfrac{5C}{2}\times {{E}^{2}}=\dfrac{5C{{E}^{2}}}{4} \\ \Rightarrow {{E}_{2}}=\dfrac{1}{2}\times \dfrac{3C}{2}\times {{E}^{2}}=\dfrac{3C{{E}^{2}}}{4} \\ \Rightarrow {{E}_{3}}=\dfrac{1}{2}\times 2C\times {{E}^{2}}=C{{E}^{2}} \\ \Rightarrow {{E}_{4}}=\dfrac{1}{2}\times C\times {{E}^{2}}=\dfrac{C{{E}^{2}}}{2} \\ \text{The total energy stored in the circuit will be;} \\ {{E}_{eq}}={{E}_{1}}+{{E}_{2}}+{{E}_{3}}+{{E}_{4}} \\ \Rightarrow {{E}_{eq}}=\left( \dfrac{5}{4}+\dfrac{3}{4}+1+\dfrac{1}{2} \right)C{{E}^{2}} \\ \therefore {{E}_{eq}}=\dfrac{7C{{E}^{2}}}{2}$

Find the total power dissipated by the circuit and the percentage contribution of the $AB$ resistor. Assume the current flowing through the $3R$ resistor to be $V$ and the current flowing through the $AB$ resistor to be $\dfrac{V}{4R}$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7ddb9606-3b4e-4606-92ae-76bb93b01a738075237682725494765.png"/>

 The total power is inversely proportional to the equivalent resistance of the circuit. 

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7b3c93f2-9791-4636-91b6-bb607707430f4866961067282511672.png"/> <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4c211774-5792-459e-b7a1-becc9b571ca51897667154148001499.png"/> $\text{The total power of the circuit is given by;} \\ {{P}_{total}}=\dfrac{{{V}^{2}}}{{{R}_{eq}}} \\ \text{From the equivalent circuit diagram;} \\ {{R}_{eq}}=\dfrac{32R}{21} \\ \text{The total power will be;} \\ \Rightarrow {{P}_{total}}=\dfrac{21{{V}^{2}}}{32R} \\ \text{The total voltage is given by;} \\ V=V\left( 3R \right)+V\left( 3R \right)+V\left( \dfrac{2R}{5} \right) \\ \text{Given that current through AB is;} \\ I=\dfrac{V}{4R} \\ \Rightarrow V=\dfrac{V}{4R}\times 3R+\dfrac{V}{4R}\times 3R+V\left( \dfrac{2R}{5} \right) \\ \Rightarrow V\left( \dfrac{2R}{5} \right)=V-\dfrac{3V}{2}=-\dfrac{V}{2} \\ \text{Here }V\left( \dfrac{2R}{5} \right)\text{ is the voltage across the parallel combination of the upper branch} \\ \text{This same potential will be across AB} \\ \text{Power dissipated by AB is given by;} \\ \Rightarrow {{P}_{AB}}=\dfrac{{{\left[ V\left( \dfrac{2R}{5} \right) \right]}^{2}}}{R}=\dfrac{{{V}^{2}}}{4R} \\ \text{Percentage contribution is given by;} \\ \Rightarrow \eta =\dfrac{{{P}_{AB}}}{{{P}_{total}}}\times 100=\dfrac{32{{V}^{2}}R}{4\times 21R{{V}^{2}}}\times 100 \\ \Rightarrow \eta =\dfrac{32}{84}\times 100 \\ \therefore \eta =38.09\%$

 A certain circuit has $n$ resistors in series, determine the factor by which the power changes if the first six resistors are connected in parallel and this parallel combination is connected in series with the remaining. 

 The power dissipated by the circuit is dependent on the arrangement of the resistors, parallel combination reduces the power and series increases the power. 

 $\text{For case one the equivalent resistance is given by;} \\ {{R}_{eq}}=nR \\ \text{Power is given by;} \\ {{P}_{1}}={{I}^{2}}{{R}_{eq}}={{I}^{2}}nR \\ \text{For the second condition the equivalent resistance is given by;} \\ {{R}_{eq}}=\dfrac{R}{6}+\left( n-6 \right)R \\ \Rightarrow {{R}_{eq}}=\left( 6n-35 \right)R \\ \text{Power is given by;} \\ {{P}_{2}}={{I}^{2}}{{R}_{eq}}={{I}^{2}}\left( 6n-35 \right)R \\ \text{The factor by which the power changes is given by;} \\ f=\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{nR}{\left( 6n-35 \right)R} \\ \therefore {{P}_{2}}=\dfrac{\left( 6n-35 \right)}{n}{{P}_{1}}$

 $n$ resistors each of resistance $R$ are connected in series such that $m$ number of resistors out of them are connected in parallel, this entire parallel combination is in series with the remaining resistors, calculate the heat dissipated if $I$ current flows through the circuit for $t$ seconds.

 The heat dissipated is directly proportional to the equivalent resistance of the circuit and the time for which the current is flowing.

 $\text{For the }m\text{ resistors connected in parallel} \\ \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+\text{ }.................\text{ }m\text{ -times} \\ \Rightarrow \dfrac{1}{{{R}_{eq}}}=\dfrac{m}{R} \\ \Rightarrow {{R}_{eq}}=\dfrac{R}{m} \\  \text{For the }\left( n-m \right)\text{ resistors connected in series;} \\ {{R}_{eq}}=R+R+R+R\text{ }..............\text{ }\left( n-m \right)\text{ times} \\ \Rightarrow \left( {} \right){{R}_{eq}}=\left( n-m \right)R \\  \text{The equivalent resistance of the circuit is given by;} \\ \Rightarrow {{R}_{eq}}=\left( n-m \right)R+\dfrac{R}{m} \\ \Rightarrow {{R}_{eq}}=\dfrac{ 1+nm-{{m}^{2}} R}{m} \\ \text{Heat dissipated is given by;} \\ H={{I}^{2}}Rt \\ \therefore H=\dfrac{{{I}^{2}}\left( 1+nm-{{m}^{2}}\right)Rt}{m}$