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

Voltage Quiz 2

physics

Find the voltage across the resistor ${{R}_{3}}$ if ${R}_{1}= {R}_{2}={R}_{3}={R}_{4}$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/52687be9-8b63-4ce7-b0ab-fd9892d20f482908512076195995361.png"/>

 The voltage across the parallel components will be the same as that of the source voltage. The voltage across the series components gets divided equally. 

The voltage across the series components gets divided equally, whereas, the voltage across the parallel components will be the same as that of the source voltage. Consider the diagram. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9a294db7-cef0-4692-9899-90dc19d383878222804751033814405.png"/>

Consider the diagram and find the amount of voltage. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/424f576b-278a-444d-b473-9867975f08348038658119930408188.png"/>

 The voltage is equal to the work done in moving the positive charge from the negative to the positive terminal. 

  $V=\dfrac{W}{q} \\ \Rightarrow V=\dfrac{4}{2} \\ \therefore V=2\,V$

Consider the figure. The maximum voltage drop will be across which bulb? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/11c76784-4c34-403d-9c6d-828bced8e7467025946217656927341.png"/>

 The resistance of the bulb is proportional to the voltage drop across the bulb. The higher the voltage drop, the maximum the power dissipated by the bulb. 

 $\text{The bulb with the maximum resistance will glow brighter}\text{. } \\ \text{The bulb with higher resistance will have greater voltage drop}\text{. } \\ \text{The brighter the bulb, the higher the power dissipation}\text{.} \\ P=\dfrac{{{V}^{2}}}{R} \\ \Rightarrow V=\sqrt{PR} \\ \therefore V\propto \sqrt{R}$

 An electric company charges Rs. $3$ per $5\,kWh$ for an electric bulb that draws a current of $6\,A$ on an average of $6$ hours a day. Compute the voltage value on which the bulb works. 

 The total energy consumption equals the product of power consumed and the time taken. The power consumed equals the product of voltage and the current drawn. 

 $\text{The total energy consumption equals power consumed by the bulb times the time taken}\text{.} \\ E=P\times t \\ \Rightarrow P=\dfrac{E}{t} \\ \text{Substitute the values in the above formula}\text{.} \\ P=\dfrac{5}{6\times 60} \\ \therefore P=\dfrac{1}{72}W \\ \text{The power consumed equals the product of voltage and current}\text{.} \\ P=V\times I \\ \Rightarrow V=\dfrac{P}{I} \\ \text{Substitute the values in the above formula}\text{.} \\ V=\dfrac{{}^{1}/{}_{72}}{6} \\ \text{Therefore, the value of voltage on which the bulb works is, } \\ \therefore V=\dfrac{1}{432}V$

The voltage value across the resistor 2 is, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cb10896c-1366-4102-abba-66a4c42953e93166175438733014312.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. 

 $\text{Only the value of current gets equally divided per every branch present in the circuit}\text{.} \\ \text{Compute the equivalent resistance}\text{.} \\ R=2+4+4 \\ \therefore R=10\Omega  \\ \text{Now compute the current through the circuit}\text{.} \\ I=\dfrac{V}{R} \\ \Rightarrow I=\dfrac{30}{10} \\ \therefore I=3\,A \\ \text{Therefore, the current through resistor 2 will be,} \\ {{I}_{2}}=\dfrac{I}{2} \\ \Rightarrow {{I}_{2}}=\dfrac{3}{2} \\ \therefore {{I}_{2}}=1.5\,A \\ \text{Thus, the voltage across the resistor 2 will be,} \\ {{V}_{2}}={{I}_{2}}{{R}_{2}} \\ \Rightarrow {{V}_{2}}=\dfrac{3}{2}\times 8 \\ \therefore {{V}_{2}}=12\,V$