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

Combination circuits Quiz 2

physics

The equivalent resistance of second and third resistors of the following circuit is, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5aca5c35-c014-4487-91b3-282f6e72a5a72764035076217875526.png"/>

 The equivalent resistance in a series combination of resistors is the sum of individual resistances. The equivalent resistance in a parallel combination of resistors is the reciprocal of the sum of individual inverse resistances. 

  $\text{The resistors 2 and 3 are in parallel connection}\text{. So, their equivalent resistance is,} \\\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \\\Rightarrow \dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{2} \\\therefore R=1\Omega$

The circuit diagrams (a) and (b) equivalent combination circuits. The value of the resistor $R$ is, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9a49a76d-0156-4637-8623-48f35cdbe8662393942583014448799.png"/>

$\text{Firstly study the circuit diagram. Compare the components placed in the diagrams (a) and (b).}\\ \text{The resistor with resistance of}\,2\Omega\,\text{is present at the same place.}\\\text{The place of the resistor with resistance of}\,5\Omega\,\text{is changed.}\\\text{So, we are left with the parallel combination of resistors}\,4\Omega.\\\text{This combination is replaced with an equivalent resistance of }\,R\Omega.\,\text{The equivalent resistance of}\,4\Omega\,\text{is}\\ \dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{4} \\\Rightarrow \dfrac{1}{R}=\dfrac{2}{4}=\dfrac{1}{2} \\\therefore R=2\Omega$

In the circuit shown below, the current flow at the point $f$ is _________than the current flow at the point $d$. &nbsp;<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5a9e76cf-5151-47f7-ae79-0ee5f581ef805375255715290996983.png"/>

 The current and resistance are inversely proportional to each other, that is, the flow of current will be least across the resistor with higher resistance and vice versa.  

 From the above diagram, it is clear that point $d$ is present at a resistor with the resistance value of $6\Omega $ and point $f$ is present at a resistor with the resistance value of $5\Omega $. So, ${{R}_{d}}>{{R}_{f}}$. As the current and resistance are inversely proportional to each other, thus, ${{I}_{f}}>{{I}_{d}}$.  

In the circuit shown below, the voltage drop at the points $b$ and $c$ is _________than the voltage drop at points $a$ and $e$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ea2eccab-cf0a-4d5a-892b-9643abed918e2599080406450660618.png"/>

 The branch with the least amount of resistance will have the minimum amount of voltage drop across it and vice versa. The voltage and resistance are directly proportional to each other, that is, the resistance will be least across the least value of voltage and vice versa. 

 From the above diagram, it is clear that the resistor present at points $b$ and $c$ has the resistance value of $5\Omega $ and points $a$ and $e$ has the resistance value of $2\Omega $. So, ${{R}_{b,c}}>{{R}_{a,e}}$. As the voltage and resistance are directly proportional to each other, thus, ${{V}_{b,c}}>{{V}_{a,e}}$.  

The voltage value across the resistor ${R_3}$ is, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/307b1602-668b-433b-aa3f-a8ec1e37f0a7135694831888514341.png"/>

 Using Kirchhoff’s current and voltage law, it becomes easy to solve the problem, but, when it comes to recognizing the type of combination circuit, the current is only the parameter that defines the type of circuit. 

 $\text{Only the value of current gets equally divided per every branch present in the circuit}\text{.} \\\text{Compute the equivalent resistance}\text{.} \\R=2+3+5 \\\therefore R=10\Omega  \\\text{Now compute the current through the circuit}\text{.} \\ I=\dfrac{V}{R} \\\Rightarrow I=\dfrac{40}{10} \\\therefore I=4\,A \\\text{Therefore, the current through resistor}\,{R_3}\,\text{will be,} \\{{I}_{3}}=\dfrac{I}{2} \\\Rightarrow {{I}_{3}}=\dfrac{4}{2} \\\therefore {{I}_{3}}=2\,A \\ \text{Thus, the voltage across the resistor}\,{R_3}\,\text{will be,} \\{{V}_{3}}={{I}_{3}}{{R}_{3}} \\\Rightarrow {{V}_{3}}=2\times 6 \\\therefore {{V}_{3}}=12\,V$