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Current Quiz 2

physics

 A cell of emf $2\,V$ is connected to resistances $1\,\Omega $ and $2\,\Omega $ in parallel.Ignore the internal resistance and calculate the current in the various parts. Suppose current through $1\,\Omega $ to be ${{I}_{1}}$; current through $2\,\Omega $ to be ${{I}_{2}}$ and current through cell to be ${{I}_{C}}$.

 Identify the different parts first and directly use Ohm’s law to determine the currents in these parts since internal resistance is given to be zero.

  Since internal resistance is zero, potential difference for the external circuit is the emf and this is the same for parallel connection. Therefore, each resistance gets 2 V. $\\ \text{Current through }1\Omega \text{ resistor, }{{I}_{1}}=\dfrac{1}{2}=0.5\text{ A} \\ \text{Current through }2\Omega \text{ resistor, }{{I}_{2}}=\dfrac{2}{2}=1\text{ A} \\ \text{Current through the cell, }{{I}_{C}}=0.5+1=1.5\text{ A}$

 A cell of emf $12\,V$ and internal resistance $0.8\,\Omega $ is connected to resistances $2\,\Omega $ and $3\,\Omega $ in parallel. Find the currents in the various branches. Suppose current through $2\,\Omega $ to be ${{I}_{1}}$; current through $3\,\Omega $ to be ${{I}_{2}}$ and current through cell to be ${{I}_{C}}$. 

Internal resistance of the cell has to be taken into account to calculate the total resistance and hence, the current through the cell. Currents through respective resistors depend upon this current through the cell. 

  Since the cell has an internal resistance, a part of the emf is used to overcome the internal resistance when a current is drawn. Potential difference for the external resistance will be less than the emf.Effective resistance of $2\,\Omega$ and $3\,\Omega$ in parallel $ \\ \dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\dfrac{2\times 3}{2+3}=1.2\Omega \\ \text{Therefore, total resistance in the circuit }=1.2+0.8=2\Omega  \\ \text{Current through the cell,} \\ {{I}_{C}}=\dfrac{\text{emf}}{\text{total resistance}} = \dfrac{12}{2} = 6\text{ A} \\ \text{Current through }2\Omega \text{ resistor,} \\ {{I}_{1}}=\dfrac{{{I}_{C}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})} = \dfrac{6\times 3}{2+3} = 3.6\text{ A} \\ \text{Current through }2\Omega \text{ resistor}, \\ {{I}_{2}}=\dfrac{{{I}_{C}}{{R}_{1}}}{({{R}_{1}}+{{R}_{2}})} = \dfrac{6\times 2}{2+3} = 2.4\text{ A}$

What happens to the brilliance of the bulbs ${{B}_{1}}$ and ${{B}_{2}}$ when the key is closed? ${{B}_{1}}$ and ${{B}_{2}}$ are identical.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2220d54d-5782-484b-a5fc-da70296deb418973254194085330999.png"/> 

 When the key is closed, parallel connection of bulb and resistance decreases the total resistance of the circuit. 

When the key is open, the same current flows through ${{B}_{1}}$ and ${{B}_{2}}$. Both glow with the same brightness. When the key is closed, ${{B}_{2}}$ and $R$ are in parallel. Hence, the total resistance decreases. A larger current flow through ${{B}_{1}}$. Clearly, the brightness of ${{B}_{1}}$ increases.Potential difference across ${{B}_{1}}$ is increased. Hence, potential difference across ${{B}_{2}}$ is reduced. Clearly, the brightness of ${{B}_{2}}$ decreases. 

 Determine the currents through each side of the triangle<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c2f9f70f-f8a1-463c-948a-6876cba152634948673944362256458.png"/> 

 From the total resistance of the circuit, total current through the cell is determined, from which, using Ohm’s law and KCL, currents in each arm of the triangle is determined. 

 From the given figure, resistances of arms $AC$ and $ABC$ are parallel and their effective resistance is given by $\\ \dfrac{2\times 4}{2+4}=\dfrac{4}{3}\Omega \\ $ Clearly, total resistance of the circuit is given by $\\ \dfrac{4}{3}+\dfrac{2}{3}=2\Omega \\ \text{Current through cell, }I=\dfrac{6}{2}=3\text{ A} \\ \text{Current through arm AC, }{{I}_{1}}=\dfrac{3\times 4}{2+4}=2\text{ A} \\ \text{Current through arm AB, }{{I}_{2}}=3-2=1\text{ A} \\ \text{Current through arm BC, }{{I}_{BC}}={{I}_{2}}=1\text{ A}$

 Suppose a current of $10\text{ A}$ splits in the ratio $2:3$ through resistors $4\,\Omega $ and $x\,\Omega $ respectively, which are connected in parallel. Calculate the value of $x$. 

When current splits into two resistors in parallel, potential differences in both resistors can be equated. 

 $\text{We have: }2p+3p=10 \\ \Rightarrow p=2 \\ {{I}_{1}}=2\times 2=4\text{ A} \\ {{I}_{2}}=3\times 2=6\text{ A} \\ \text{Further, we know that} \\ {{I}_{1}}{{R}_{1}}={{I}_{2}}{{R}_{2}} \\ \Rightarrow 4\times 4=6x \\ \therefore x=\dfrac{16}{6}=\dfrac{8}{3} \\ $