151991
Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \mathrm{~V}$ battery is connected. The current through $3 \Omega$ resistor is :
1 $0.75 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Parallel combination- $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{3}+\frac{1}{3}$ $=\frac{2}{3}$ $\mathrm{R}_{\mathrm{eq}} =\frac{3}{2} \Omega$ $\mathrm{I}_{\mathrm{eq}}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{3} \times 2=2 \mathrm{~A}$ $\because$ The current through $3 \Omega$ resister, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{3}}=\frac{3}{3}=1 \mathrm{~A}$
Karnataka CET-2009
Current Electricity
151992
The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$ )
1 $30 \Omega$
2 $8 \Omega$
3 $10 \Omega$
4 $40 \Omega$
Explanation:
B The equivalent resistance between $\mathrm{D}$ and $\mathrm{C}$, $\mathrm{R}_{\mathrm{DC}}=\frac{15 \times(15+15)}{15+(15+15)}=10 \Omega$ Now, between $\mathrm{A}$ and $\mathrm{B}$ the resistance of upper part ADCB, $\mathrm{R}_{1}=15+10+15=40 \Omega$ Resistance between $\mathrm{A}$ and $\mathrm{B}$ of middle part $\mathrm{AOB}$. $\mathrm{R}_{2}=15+15=30 \Omega$ Therefore, equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}^{\prime}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15}$ $=\frac{3+4+8}{120}=\frac{15}{120}$ $\mathrm{R}^{\prime} =\frac{120}{15}=8 \Omega$
Karnataka CET-2008
Current Electricity
151993
The quantity of a charge that will be transferred by a current flow of 20 A over $1 \mathrm{~h}$ $30 \mathrm{~min}$ period is :
1 $10.8 \times 10^{3} \mathrm{C}$
2 $10.8 \times 10^{4} \mathrm{C}$
3 $5.4 \times 10^{3} \mathrm{C}$
4 $1.8 \times 10^{4} \mathrm{C}$
Explanation:
B Given, Current flow $(\mathrm{I})=20 \mathrm{~A}$ Time period $(\mathrm{t})=1 \mathrm{hr} 30 \mathrm{~min}=5400 \mathrm{sec}$. Quantity of charge transferred $(\mathrm{Q})=\mathrm{It}$ $=20 \times 5400$ $=10.8 \times 10^{4} \mathrm{C}$
151991
Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \mathrm{~V}$ battery is connected. The current through $3 \Omega$ resistor is :
1 $0.75 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Parallel combination- $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{3}+\frac{1}{3}$ $=\frac{2}{3}$ $\mathrm{R}_{\mathrm{eq}} =\frac{3}{2} \Omega$ $\mathrm{I}_{\mathrm{eq}}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{3} \times 2=2 \mathrm{~A}$ $\because$ The current through $3 \Omega$ resister, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{3}}=\frac{3}{3}=1 \mathrm{~A}$
Karnataka CET-2009
Current Electricity
151992
The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$ )
1 $30 \Omega$
2 $8 \Omega$
3 $10 \Omega$
4 $40 \Omega$
Explanation:
B The equivalent resistance between $\mathrm{D}$ and $\mathrm{C}$, $\mathrm{R}_{\mathrm{DC}}=\frac{15 \times(15+15)}{15+(15+15)}=10 \Omega$ Now, between $\mathrm{A}$ and $\mathrm{B}$ the resistance of upper part ADCB, $\mathrm{R}_{1}=15+10+15=40 \Omega$ Resistance between $\mathrm{A}$ and $\mathrm{B}$ of middle part $\mathrm{AOB}$. $\mathrm{R}_{2}=15+15=30 \Omega$ Therefore, equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}^{\prime}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15}$ $=\frac{3+4+8}{120}=\frac{15}{120}$ $\mathrm{R}^{\prime} =\frac{120}{15}=8 \Omega$
Karnataka CET-2008
Current Electricity
151993
The quantity of a charge that will be transferred by a current flow of 20 A over $1 \mathrm{~h}$ $30 \mathrm{~min}$ period is :
1 $10.8 \times 10^{3} \mathrm{C}$
2 $10.8 \times 10^{4} \mathrm{C}$
3 $5.4 \times 10^{3} \mathrm{C}$
4 $1.8 \times 10^{4} \mathrm{C}$
Explanation:
B Given, Current flow $(\mathrm{I})=20 \mathrm{~A}$ Time period $(\mathrm{t})=1 \mathrm{hr} 30 \mathrm{~min}=5400 \mathrm{sec}$. Quantity of charge transferred $(\mathrm{Q})=\mathrm{It}$ $=20 \times 5400$ $=10.8 \times 10^{4} \mathrm{C}$
151991
Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \mathrm{~V}$ battery is connected. The current through $3 \Omega$ resistor is :
1 $0.75 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Parallel combination- $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{3}+\frac{1}{3}$ $=\frac{2}{3}$ $\mathrm{R}_{\mathrm{eq}} =\frac{3}{2} \Omega$ $\mathrm{I}_{\mathrm{eq}}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{3} \times 2=2 \mathrm{~A}$ $\because$ The current through $3 \Omega$ resister, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{3}}=\frac{3}{3}=1 \mathrm{~A}$
Karnataka CET-2009
Current Electricity
151992
The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$ )
1 $30 \Omega$
2 $8 \Omega$
3 $10 \Omega$
4 $40 \Omega$
Explanation:
B The equivalent resistance between $\mathrm{D}$ and $\mathrm{C}$, $\mathrm{R}_{\mathrm{DC}}=\frac{15 \times(15+15)}{15+(15+15)}=10 \Omega$ Now, between $\mathrm{A}$ and $\mathrm{B}$ the resistance of upper part ADCB, $\mathrm{R}_{1}=15+10+15=40 \Omega$ Resistance between $\mathrm{A}$ and $\mathrm{B}$ of middle part $\mathrm{AOB}$. $\mathrm{R}_{2}=15+15=30 \Omega$ Therefore, equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}^{\prime}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15}$ $=\frac{3+4+8}{120}=\frac{15}{120}$ $\mathrm{R}^{\prime} =\frac{120}{15}=8 \Omega$
Karnataka CET-2008
Current Electricity
151993
The quantity of a charge that will be transferred by a current flow of 20 A over $1 \mathrm{~h}$ $30 \mathrm{~min}$ period is :
1 $10.8 \times 10^{3} \mathrm{C}$
2 $10.8 \times 10^{4} \mathrm{C}$
3 $5.4 \times 10^{3} \mathrm{C}$
4 $1.8 \times 10^{4} \mathrm{C}$
Explanation:
B Given, Current flow $(\mathrm{I})=20 \mathrm{~A}$ Time period $(\mathrm{t})=1 \mathrm{hr} 30 \mathrm{~min}=5400 \mathrm{sec}$. Quantity of charge transferred $(\mathrm{Q})=\mathrm{It}$ $=20 \times 5400$ $=10.8 \times 10^{4} \mathrm{C}$
151991
Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \mathrm{~V}$ battery is connected. The current through $3 \Omega$ resistor is :
1 $0.75 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Parallel combination- $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{3}+\frac{1}{3}$ $=\frac{2}{3}$ $\mathrm{R}_{\mathrm{eq}} =\frac{3}{2} \Omega$ $\mathrm{I}_{\mathrm{eq}}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{3} \times 2=2 \mathrm{~A}$ $\because$ The current through $3 \Omega$ resister, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{3}}=\frac{3}{3}=1 \mathrm{~A}$
Karnataka CET-2009
Current Electricity
151992
The equivalent resistance between the points $A$ and $B$ will be (each resistance is $15 \Omega$ )
1 $30 \Omega$
2 $8 \Omega$
3 $10 \Omega$
4 $40 \Omega$
Explanation:
B The equivalent resistance between $\mathrm{D}$ and $\mathrm{C}$, $\mathrm{R}_{\mathrm{DC}}=\frac{15 \times(15+15)}{15+(15+15)}=10 \Omega$ Now, between $\mathrm{A}$ and $\mathrm{B}$ the resistance of upper part ADCB, $\mathrm{R}_{1}=15+10+15=40 \Omega$ Resistance between $\mathrm{A}$ and $\mathrm{B}$ of middle part $\mathrm{AOB}$. $\mathrm{R}_{2}=15+15=30 \Omega$ Therefore, equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}^{\prime}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15}$ $=\frac{3+4+8}{120}=\frac{15}{120}$ $\mathrm{R}^{\prime} =\frac{120}{15}=8 \Omega$
Karnataka CET-2008
Current Electricity
151993
The quantity of a charge that will be transferred by a current flow of 20 A over $1 \mathrm{~h}$ $30 \mathrm{~min}$ period is :
1 $10.8 \times 10^{3} \mathrm{C}$
2 $10.8 \times 10^{4} \mathrm{C}$
3 $5.4 \times 10^{3} \mathrm{C}$
4 $1.8 \times 10^{4} \mathrm{C}$
Explanation:
B Given, Current flow $(\mathrm{I})=20 \mathrm{~A}$ Time period $(\mathrm{t})=1 \mathrm{hr} 30 \mathrm{~min}=5400 \mathrm{sec}$. Quantity of charge transferred $(\mathrm{Q})=\mathrm{It}$ $=20 \times 5400$ $=10.8 \times 10^{4} \mathrm{C}$
