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This is second of our many quizzes on Series circuits. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Series circuits Quiz 2

physics

Consider the series circuit below, if a single resistance of $18\Omega $ offers the same resistance as the two resistance ${{R}_{1}}$ and ${{R}_{2}}$ shown below, then what is the value of ${{R}_{2}}$ if ${{R}_{1}}=8\Omega $?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6ab058b7-59a3-42e8-94e6-783d209041232018375362817289867.png"/> 

When two resistances are in series the current flowing through resistors ${{R}_{1}}$ and ${{R}_{2}}$ is same and voltage drop across ${{R}_{1}}$ and ${{R}_{2}}$ is different. So, the equivalent resistance is given by ${{R}_{eq}}={{R}_{1}}+{{R}_{2}}$.Put the values given in the question to calculate the unknown resistance.

 $\text{In the figure given in the question, two resistances are in series and connected to a battery of emf E,}\\ \text{the current flowing through resistors}\,{{R}_{1}}\,\text{and}\,{{R}_{2}}\,\text{is same}\\ \text{and voltage drop across}\,{{R}_{1}}\,\text{and}\,{{R}_{2}}\,\text{is different.}\\ \text{So, the equivalent resistance is given by,}\\ {{R}_{eq}}={{R}_{1}}+{{R}_{2}}\\ \text{According to the question,}\\ {{R}_{1}}=8\Omega ,{{R}_{eq}}=18\Omega\\ \text{Putting the above values in the equation we get,}\\ \therefore{{R}_{2}}=10\Omega $ 

A refrigerator and freezer are built using,

 In a series circuit, the components are attached in a single line such that the current has only one pathway, the refrigerators and freezers both use a series circuit. 

 Refrigerators and freezers both use series circuits. As we know, in a series circuit, the components are attached in a single line such that the current has only one pathway. If there is a break in the circuit then all components in the circuit are turned off simultaneously. The compressor and the temperature control switch which senses any drop in the temperature of the refrigerator, they are all connected in series. So, in case of drop in temperature, the control switch is turned on and all the components are initiated until the temperature decreases as per requirements. 

Consider the series $RC$ circuit, if total impedance is $19.7\Omega $, then what is the value of the resistance(R) in series? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f1023bc8-29e8-4679-8bf1-f0229f9367001048739267682348204.png"/>

The total impedance $Z$, in a series $RC$ circuit is given by $Z=\sqrt{{{R}^{2}}+{{(\dfrac{1}{\omega C})}^{2}}}$.To calculate the resistance of the resistor, we put the values given in the question in the above equation, since all other variables are known we solve for $R$.

$\text{The total impedance in a series RC circuit is given by,}\\ Z=\sqrt{{{R}^{2}}+{{(\dfrac{1}{\omega C})}^{2}}}….(i)\\ \text{According to the question,}\\ C=50\mu F,\omega =50Hz,Z=19.7\Omega\\ \text{Putting the above values in equation (i), and solving for R, we get,}\\ \therefore R=34\Omega $ 

Consider the circuit shown below, the total impedance of this circuit will be,<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cee77d85-acb9-4479-a197-528ea8f8e9c68491539650802435674.png"/>

The total impedance of $LCR$ series circuit is given by $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$.Put the values given in question to find the total impedance of the circuit. 

$\text{The circuit in question is a series LCR circuit, with a resistance R, capacitance C}\\ \text{and an inductor L, all connected in series.The total impedance of LCR series circuit is given by,}\\ Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}….(i)\\ \text{According to the question,}\\ R=12\Omega ,C=50\mu F,L=40H,f=50Hz\\ \text{Putting the above values in equation (i) and solving, we get,}\\ Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=\sqrt{{{R}^{2}}+{{(2\pi fL-\dfrac{1}{2\pi fC})}^{2}}} \\ \Rightarrow Z=\sqrt{{{12}^{2}}+{{(2\pi \times 50\times 40-\dfrac{1}{2\pi \times 50\times 50\times {{10}^{-6}}})}^{2}}}=12.5k\Omega  \\ \therefore Z=12.5k\Omega$ 

 In the $RL$ series shown below, if the total impedance of the circuit is $176.2\,\Omega $, then what is the value of $L$ and the voltage drop across the inductor?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/38bf4607-2c1e-45e9-9e62-13f32c0b82628267547396359570741.png"/>

The total impedance of $RL$ series circuit and voltage drop across the inductor is given by,$Z=\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}=\sqrt{{{R}^{2}}+{{(2\pi fL)}^{2}}}$ and ${{V}_{L}}=I{{X}_{L}}=I\times 2\pi fL$. Put the values and find the unknowns. 

$\text{In the RL series, with a resistance R and an inductor L are connected in series.}\\ \text{The total impedance of RL series circuit is given by,}\\ Z=\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}=\sqrt{{{R}^{2}}+{{(2\pi fL)}^{2}}}….(i)\\ \text{According to the question,}\\ Z=176.2\Omega ,R=12\Omega ,f=50Hz\,\text{and current in the circuit I=5A}\\ \text{Putting the above values in equation (i) and solving, we get,}\\ Z=\sqrt{{{R}^{2}}+{{(2\pi fL)}^{2}}} \\ \Rightarrow 176.2\Omega =\sqrt{{{12}^{2}}+{{(2\pi \times 50\times L)}^{2}}} \\ \Rightarrow L=0.56H \\ \text{And the voltage drop across the inductor is given by,}\\{{V}_{L}}=I{{X}_{L}}\\ \Rightarrow{{V}_{L}} =I\times 2\pi fL\\ \Rightarrow{{V}_{L}} =5\times 2\pi \times 50\times 0.56\\ \therefore{{V}_{L}} =0.88mV$ 