151881
Find the equivalent resistance across $A B$
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
Explanation:
A $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=\frac{4}{4}=1$ $\mathrm{R}_{\mathrm{AB}}=1 \Omega$
Manipal UGET-2019
Current Electricity
151883
The graphs between current (I) and voltage (V) for three linear resistors 1,2 and 3 are given below: If $R_{1}, R_{2}$ and $R_{3}$ are the resistances at these resistors, then which one of the following is correct?
1 $R_{1}>R_{2}>R_{3}$
2 $R_{1}\ltR_{3}\ltR_{2}$
3 $R_{3}\ltR_{1}\ltR_{2}$
4 $R_{3}>R_{2}>R_{1}$
Explanation:
B By Ohm's law, $\mathrm{V} \propto \mathrm{I}$ $\mathrm{V}=\mathrm{IR}$ $\frac{\mathrm{V}}{\mathrm{I}}=\tan \theta=\mathrm{R}$ $(\tan \theta \uparrow, \theta \uparrow)$ The graph between current and voltage is $\theta_{1}\lt\theta_{3}\lt\theta_{2}$ $\tan \theta_{1}\lt\tan \theta_{3}\lt\tan \theta_{2}$ $\mathrm{R}_{1}\lt\mathrm{R}_{3}\lt\mathrm{R}_{2}$
NDA (II) 2018
Current Electricity
151884
Find $i$ in shown figure
1 $0.2 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $0.4 \mathrm{~A}$
Explanation:
B Here, $30 \Omega$ and $60 \Omega$ resistors are in parallel. Let's assume ; $\mathrm{R}_{1}=30 \Omega, \mathrm{R}_{2}=60 \Omega$ Then, $\mathrm{R}_{\text {net }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{30 \times 60}{30+60}=20 \Omega$ We know that, $i=\frac{V}{R}$ $\therefore \quad i=\frac{V}{R_{\text {net }}}=\frac{2}{20}=\frac{1}{10}=0.1 \mathrm{~A}$
JIPMER-2018
Current Electricity
151885
Find current (i) in circuit shown in figure.
1 $0.5 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A $\because \quad \frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}}$ The figure satisfies Wheat stone bridge. Hence, there is no current flowing though BC. Therefore, the circuit becomes- $\mathrm{R}_{1}=\mathrm{P}+\mathrm{Q}=5+10=15 \Omega$ $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}=10+20=30 \Omega$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{15 \times 30}{15+30}=\frac{450}{45}=10 \Omega$ $\because \quad$ Current, $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $\mathrm{i}=\frac{5}{10}=0.5 \mathrm{~A}$
151881
Find the equivalent resistance across $A B$
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
Explanation:
A $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=\frac{4}{4}=1$ $\mathrm{R}_{\mathrm{AB}}=1 \Omega$
Manipal UGET-2019
Current Electricity
151883
The graphs between current (I) and voltage (V) for three linear resistors 1,2 and 3 are given below: If $R_{1}, R_{2}$ and $R_{3}$ are the resistances at these resistors, then which one of the following is correct?
1 $R_{1}>R_{2}>R_{3}$
2 $R_{1}\ltR_{3}\ltR_{2}$
3 $R_{3}\ltR_{1}\ltR_{2}$
4 $R_{3}>R_{2}>R_{1}$
Explanation:
B By Ohm's law, $\mathrm{V} \propto \mathrm{I}$ $\mathrm{V}=\mathrm{IR}$ $\frac{\mathrm{V}}{\mathrm{I}}=\tan \theta=\mathrm{R}$ $(\tan \theta \uparrow, \theta \uparrow)$ The graph between current and voltage is $\theta_{1}\lt\theta_{3}\lt\theta_{2}$ $\tan \theta_{1}\lt\tan \theta_{3}\lt\tan \theta_{2}$ $\mathrm{R}_{1}\lt\mathrm{R}_{3}\lt\mathrm{R}_{2}$
NDA (II) 2018
Current Electricity
151884
Find $i$ in shown figure
1 $0.2 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $0.4 \mathrm{~A}$
Explanation:
B Here, $30 \Omega$ and $60 \Omega$ resistors are in parallel. Let's assume ; $\mathrm{R}_{1}=30 \Omega, \mathrm{R}_{2}=60 \Omega$ Then, $\mathrm{R}_{\text {net }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{30 \times 60}{30+60}=20 \Omega$ We know that, $i=\frac{V}{R}$ $\therefore \quad i=\frac{V}{R_{\text {net }}}=\frac{2}{20}=\frac{1}{10}=0.1 \mathrm{~A}$
JIPMER-2018
Current Electricity
151885
Find current (i) in circuit shown in figure.
1 $0.5 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A $\because \quad \frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}}$ The figure satisfies Wheat stone bridge. Hence, there is no current flowing though BC. Therefore, the circuit becomes- $\mathrm{R}_{1}=\mathrm{P}+\mathrm{Q}=5+10=15 \Omega$ $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}=10+20=30 \Omega$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{15 \times 30}{15+30}=\frac{450}{45}=10 \Omega$ $\because \quad$ Current, $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $\mathrm{i}=\frac{5}{10}=0.5 \mathrm{~A}$
151881
Find the equivalent resistance across $A B$
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
Explanation:
A $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=\frac{4}{4}=1$ $\mathrm{R}_{\mathrm{AB}}=1 \Omega$
Manipal UGET-2019
Current Electricity
151883
The graphs between current (I) and voltage (V) for three linear resistors 1,2 and 3 are given below: If $R_{1}, R_{2}$ and $R_{3}$ are the resistances at these resistors, then which one of the following is correct?
1 $R_{1}>R_{2}>R_{3}$
2 $R_{1}\ltR_{3}\ltR_{2}$
3 $R_{3}\ltR_{1}\ltR_{2}$
4 $R_{3}>R_{2}>R_{1}$
Explanation:
B By Ohm's law, $\mathrm{V} \propto \mathrm{I}$ $\mathrm{V}=\mathrm{IR}$ $\frac{\mathrm{V}}{\mathrm{I}}=\tan \theta=\mathrm{R}$ $(\tan \theta \uparrow, \theta \uparrow)$ The graph between current and voltage is $\theta_{1}\lt\theta_{3}\lt\theta_{2}$ $\tan \theta_{1}\lt\tan \theta_{3}\lt\tan \theta_{2}$ $\mathrm{R}_{1}\lt\mathrm{R}_{3}\lt\mathrm{R}_{2}$
NDA (II) 2018
Current Electricity
151884
Find $i$ in shown figure
1 $0.2 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $0.4 \mathrm{~A}$
Explanation:
B Here, $30 \Omega$ and $60 \Omega$ resistors are in parallel. Let's assume ; $\mathrm{R}_{1}=30 \Omega, \mathrm{R}_{2}=60 \Omega$ Then, $\mathrm{R}_{\text {net }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{30 \times 60}{30+60}=20 \Omega$ We know that, $i=\frac{V}{R}$ $\therefore \quad i=\frac{V}{R_{\text {net }}}=\frac{2}{20}=\frac{1}{10}=0.1 \mathrm{~A}$
JIPMER-2018
Current Electricity
151885
Find current (i) in circuit shown in figure.
1 $0.5 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A $\because \quad \frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}}$ The figure satisfies Wheat stone bridge. Hence, there is no current flowing though BC. Therefore, the circuit becomes- $\mathrm{R}_{1}=\mathrm{P}+\mathrm{Q}=5+10=15 \Omega$ $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}=10+20=30 \Omega$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{15 \times 30}{15+30}=\frac{450}{45}=10 \Omega$ $\because \quad$ Current, $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $\mathrm{i}=\frac{5}{10}=0.5 \mathrm{~A}$
151881
Find the equivalent resistance across $A B$
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
Explanation:
A $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=\frac{4}{4}=1$ $\mathrm{R}_{\mathrm{AB}}=1 \Omega$
Manipal UGET-2019
Current Electricity
151883
The graphs between current (I) and voltage (V) for three linear resistors 1,2 and 3 are given below: If $R_{1}, R_{2}$ and $R_{3}$ are the resistances at these resistors, then which one of the following is correct?
1 $R_{1}>R_{2}>R_{3}$
2 $R_{1}\ltR_{3}\ltR_{2}$
3 $R_{3}\ltR_{1}\ltR_{2}$
4 $R_{3}>R_{2}>R_{1}$
Explanation:
B By Ohm's law, $\mathrm{V} \propto \mathrm{I}$ $\mathrm{V}=\mathrm{IR}$ $\frac{\mathrm{V}}{\mathrm{I}}=\tan \theta=\mathrm{R}$ $(\tan \theta \uparrow, \theta \uparrow)$ The graph between current and voltage is $\theta_{1}\lt\theta_{3}\lt\theta_{2}$ $\tan \theta_{1}\lt\tan \theta_{3}\lt\tan \theta_{2}$ $\mathrm{R}_{1}\lt\mathrm{R}_{3}\lt\mathrm{R}_{2}$
NDA (II) 2018
Current Electricity
151884
Find $i$ in shown figure
1 $0.2 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $0.4 \mathrm{~A}$
Explanation:
B Here, $30 \Omega$ and $60 \Omega$ resistors are in parallel. Let's assume ; $\mathrm{R}_{1}=30 \Omega, \mathrm{R}_{2}=60 \Omega$ Then, $\mathrm{R}_{\text {net }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{30 \times 60}{30+60}=20 \Omega$ We know that, $i=\frac{V}{R}$ $\therefore \quad i=\frac{V}{R_{\text {net }}}=\frac{2}{20}=\frac{1}{10}=0.1 \mathrm{~A}$
JIPMER-2018
Current Electricity
151885
Find current (i) in circuit shown in figure.
1 $0.5 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A $\because \quad \frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}}$ The figure satisfies Wheat stone bridge. Hence, there is no current flowing though BC. Therefore, the circuit becomes- $\mathrm{R}_{1}=\mathrm{P}+\mathrm{Q}=5+10=15 \Omega$ $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}=10+20=30 \Omega$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{15 \times 30}{15+30}=\frac{450}{45}=10 \Omega$ $\because \quad$ Current, $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$ $\mathrm{i}=\frac{5}{10}=0.5 \mathrm{~A}$
