151864
Masses of three wires of copper are in the ratio $1: 3: 5$ and their lengths are in the ratio $5: 3$ : 1. The ratio of their electrical resistance are :
1 $1: 3: 5$
2 $5: 3: 1$
3 $1: 15: 125$
4 $125: 15: 1$
Explanation:
D Given, Ratio of mass of wire $\mathrm{m}_{1}: \mathrm{m}_{2}: \mathrm{m}_{3}=1: 3: 5$ Ratio of length of wire $l_{1}: l_{2}: l_{3}=5: 3: 1$ We know that - $\text { Resistance }(\mathrm{R})=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}} \times \frac{l}{l}$ $\mathrm{R}=\rho \frac{l^{2}}{\mathrm{~V}}$ ${[\because \mathrm{A} \times l=\mathrm{V}]}$ $\mathrm{R}=\rho \frac{l^{2}}{\frac{\mathrm{m}}{\mathrm{d}}}$ \(\begin{aligned} \mathrm{R}=\rho \frac{l^2}{\mathrm{~V}} & {[\because \mathrm{A} \times l=\mathrm{V}] } \\ \mathrm{R}=\rho \frac{l^2}{\frac{\mathrm{m}}{\mathrm{d}}} & \left\{\begin{aligned} & \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \\ & \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\end{aligned}\right\}\end{aligned}\) Where, $\rho \rightarrow$ Resistivity of wire $\mathrm{d} \rightarrow \text { density of wire }$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}}$ So, $\quad \mathrm{R}_{1} \propto \frac{l_{1}^{2}}{\mathrm{~m}_{1}}, \mathrm{R}_{2} \propto \frac{l_{2}^{2}}{\mathrm{~m}_{2}}$ and $\mathrm{R}_{3} \propto \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =\frac{l_{1}^{2}}{\mathrm{~m}_{1}}: \frac{l_{2}^{2}}{\mathrm{~m}_{2}}: \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}=25: 3: \frac{1}{5}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =125: 15: 1$
Karnataka CET-2019
Current Electricity
151866
The value of current (in Ampere) through the 3 $\Omega$ resistance in the circuit as shown is
1 $0.05 \mathrm{~A}$
2 $0.25 \mathrm{~A}$
3 $0.76 \mathrm{~A}$
4 $0.96 \mathrm{~A}$
Explanation:
D From figure, it is clear that $0.1 \Omega$ and $60 \Omega$ resistor are in parallel. So, $\frac{1}{\mathrm{R}} =\frac{1}{0.1}+\frac{1}{60}$ $=10+\frac{1}{60}$ $\frac{1}{\mathrm{R}} =\frac{601}{60}$ $\mathrm{R} =\frac{60}{601} \Omega$ Now, $R \Omega$ and $3 \Omega$ are in series combination. Hence, $\mathrm{R}^{\prime} =\mathrm{R}+3$ $=\frac{60}{601}+3=\frac{60+1803}{601}$ $\mathrm{R}^{\prime} =\frac{1863}{601}$ Current through $3 \Omega$ resistance, $i=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $=\frac{3}{\frac{1863}{601}}=\frac{1803}{1863}$ $\mathrm{i}=0.96 \mathrm{~A}$
AMU-2019
Current Electricity
151867
In the given circuit, find voltage across $12 \Omega$ resistance.
1 12 Volt
2 36 Volt
3 72 Volt
4 48 Volt
Explanation:
D After simplifying the circuit we get, $\mathrm{R}_{\text {eq }}=\frac{6 \times 12}{6+12}=4 \Omega$ $\mathrm{I}=12 \mathrm{~A}$ $\mathrm{~V}=\mathrm{IR}$ $\quad=12 \times 4$ $\mathrm{V}=48 \text { Volt. }$
JIPMER-2019
Current Electricity
151868
A circuit contain two resistance $R_{1}$ and $R_{2}$ in series. Find the ratio of input voltage to voltage of $\mathbf{R}_{\mathbf{2}}$.
B From figure, $\text { Input voltage, } \mathrm{V}=\mathrm{i}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)$ Voltage across $R_{2}, V_{R_{2}}=i R_{2}$ Required ratio $=\frac{\text { Input voltage }}{\text { voltage across } \mathrm{R}_{2}}$ $=\frac{V}{V_{R_{2}}}=\frac{i\left(R_{1}+R_{2}\right)}{i R_{2}}$ $=\frac{R_{1}+R_{2}}{R_{2}}$
JIPMER-2019
Current Electricity
151869
An electric cable has just one copper wire of radius $8 \mathrm{~mm}$ and resistance $10 \Omega$. This single copper wire cable is replaced by 4 different well insulated copper wires each of radius 6 $\mathrm{mm}$, then total resistance of the cable is
1 $4.44 \Omega$
2 $3.33 \Omega$
3 $2.22 \Omega$
4 $5.55 \Omega$
Explanation:
A Given, $\mathrm{r}_{1}=8 \mathrm{~mm}, \mathrm{r}_{2}=6 \mathrm{~mm}, \mathrm{R}_{1}=10 \Omega$ As we know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ Where, $\mathrm{R}=$ Resistance of wire $l=\text { length of wire }$ $\mathrm{A}=$ cross section of wire $\because \quad \mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\frac{10}{\mathrm{R}_{2}}=\frac{6^{2}}{8^{2}}=\frac{36}{64}$ $\frac{10}{\mathrm{R}_{2}}=\frac{36}{64}$ $\mathrm{R}_{2}=\frac{640}{36}$ $\mathrm{R}_{2}=\frac{160}{9}$ Equivalent resistance of 4 wire each of resistance $R_{2}$ connected in parallel is $\mathrm{R}^{\prime}$ $\mathrm{R}^{\prime}=\frac{\mathrm{R}_{2}}{4}=\frac{160}{9 \times 4}=4.44 \Omega$
151864
Masses of three wires of copper are in the ratio $1: 3: 5$ and their lengths are in the ratio $5: 3$ : 1. The ratio of their electrical resistance are :
1 $1: 3: 5$
2 $5: 3: 1$
3 $1: 15: 125$
4 $125: 15: 1$
Explanation:
D Given, Ratio of mass of wire $\mathrm{m}_{1}: \mathrm{m}_{2}: \mathrm{m}_{3}=1: 3: 5$ Ratio of length of wire $l_{1}: l_{2}: l_{3}=5: 3: 1$ We know that - $\text { Resistance }(\mathrm{R})=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}} \times \frac{l}{l}$ $\mathrm{R}=\rho \frac{l^{2}}{\mathrm{~V}}$ ${[\because \mathrm{A} \times l=\mathrm{V}]}$ $\mathrm{R}=\rho \frac{l^{2}}{\frac{\mathrm{m}}{\mathrm{d}}}$ \(\begin{aligned} \mathrm{R}=\rho \frac{l^2}{\mathrm{~V}} & {[\because \mathrm{A} \times l=\mathrm{V}] } \\ \mathrm{R}=\rho \frac{l^2}{\frac{\mathrm{m}}{\mathrm{d}}} & \left\{\begin{aligned} & \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \\ & \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\end{aligned}\right\}\end{aligned}\) Where, $\rho \rightarrow$ Resistivity of wire $\mathrm{d} \rightarrow \text { density of wire }$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}}$ So, $\quad \mathrm{R}_{1} \propto \frac{l_{1}^{2}}{\mathrm{~m}_{1}}, \mathrm{R}_{2} \propto \frac{l_{2}^{2}}{\mathrm{~m}_{2}}$ and $\mathrm{R}_{3} \propto \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =\frac{l_{1}^{2}}{\mathrm{~m}_{1}}: \frac{l_{2}^{2}}{\mathrm{~m}_{2}}: \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}=25: 3: \frac{1}{5}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =125: 15: 1$
Karnataka CET-2019
Current Electricity
151866
The value of current (in Ampere) through the 3 $\Omega$ resistance in the circuit as shown is
1 $0.05 \mathrm{~A}$
2 $0.25 \mathrm{~A}$
3 $0.76 \mathrm{~A}$
4 $0.96 \mathrm{~A}$
Explanation:
D From figure, it is clear that $0.1 \Omega$ and $60 \Omega$ resistor are in parallel. So, $\frac{1}{\mathrm{R}} =\frac{1}{0.1}+\frac{1}{60}$ $=10+\frac{1}{60}$ $\frac{1}{\mathrm{R}} =\frac{601}{60}$ $\mathrm{R} =\frac{60}{601} \Omega$ Now, $R \Omega$ and $3 \Omega$ are in series combination. Hence, $\mathrm{R}^{\prime} =\mathrm{R}+3$ $=\frac{60}{601}+3=\frac{60+1803}{601}$ $\mathrm{R}^{\prime} =\frac{1863}{601}$ Current through $3 \Omega$ resistance, $i=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $=\frac{3}{\frac{1863}{601}}=\frac{1803}{1863}$ $\mathrm{i}=0.96 \mathrm{~A}$
AMU-2019
Current Electricity
151867
In the given circuit, find voltage across $12 \Omega$ resistance.
1 12 Volt
2 36 Volt
3 72 Volt
4 48 Volt
Explanation:
D After simplifying the circuit we get, $\mathrm{R}_{\text {eq }}=\frac{6 \times 12}{6+12}=4 \Omega$ $\mathrm{I}=12 \mathrm{~A}$ $\mathrm{~V}=\mathrm{IR}$ $\quad=12 \times 4$ $\mathrm{V}=48 \text { Volt. }$
JIPMER-2019
Current Electricity
151868
A circuit contain two resistance $R_{1}$ and $R_{2}$ in series. Find the ratio of input voltage to voltage of $\mathbf{R}_{\mathbf{2}}$.
B From figure, $\text { Input voltage, } \mathrm{V}=\mathrm{i}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)$ Voltage across $R_{2}, V_{R_{2}}=i R_{2}$ Required ratio $=\frac{\text { Input voltage }}{\text { voltage across } \mathrm{R}_{2}}$ $=\frac{V}{V_{R_{2}}}=\frac{i\left(R_{1}+R_{2}\right)}{i R_{2}}$ $=\frac{R_{1}+R_{2}}{R_{2}}$
JIPMER-2019
Current Electricity
151869
An electric cable has just one copper wire of radius $8 \mathrm{~mm}$ and resistance $10 \Omega$. This single copper wire cable is replaced by 4 different well insulated copper wires each of radius 6 $\mathrm{mm}$, then total resistance of the cable is
1 $4.44 \Omega$
2 $3.33 \Omega$
3 $2.22 \Omega$
4 $5.55 \Omega$
Explanation:
A Given, $\mathrm{r}_{1}=8 \mathrm{~mm}, \mathrm{r}_{2}=6 \mathrm{~mm}, \mathrm{R}_{1}=10 \Omega$ As we know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ Where, $\mathrm{R}=$ Resistance of wire $l=\text { length of wire }$ $\mathrm{A}=$ cross section of wire $\because \quad \mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\frac{10}{\mathrm{R}_{2}}=\frac{6^{2}}{8^{2}}=\frac{36}{64}$ $\frac{10}{\mathrm{R}_{2}}=\frac{36}{64}$ $\mathrm{R}_{2}=\frac{640}{36}$ $\mathrm{R}_{2}=\frac{160}{9}$ Equivalent resistance of 4 wire each of resistance $R_{2}$ connected in parallel is $\mathrm{R}^{\prime}$ $\mathrm{R}^{\prime}=\frac{\mathrm{R}_{2}}{4}=\frac{160}{9 \times 4}=4.44 \Omega$
151864
Masses of three wires of copper are in the ratio $1: 3: 5$ and their lengths are in the ratio $5: 3$ : 1. The ratio of their electrical resistance are :
1 $1: 3: 5$
2 $5: 3: 1$
3 $1: 15: 125$
4 $125: 15: 1$
Explanation:
D Given, Ratio of mass of wire $\mathrm{m}_{1}: \mathrm{m}_{2}: \mathrm{m}_{3}=1: 3: 5$ Ratio of length of wire $l_{1}: l_{2}: l_{3}=5: 3: 1$ We know that - $\text { Resistance }(\mathrm{R})=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}} \times \frac{l}{l}$ $\mathrm{R}=\rho \frac{l^{2}}{\mathrm{~V}}$ ${[\because \mathrm{A} \times l=\mathrm{V}]}$ $\mathrm{R}=\rho \frac{l^{2}}{\frac{\mathrm{m}}{\mathrm{d}}}$ \(\begin{aligned} \mathrm{R}=\rho \frac{l^2}{\mathrm{~V}} & {[\because \mathrm{A} \times l=\mathrm{V}] } \\ \mathrm{R}=\rho \frac{l^2}{\frac{\mathrm{m}}{\mathrm{d}}} & \left\{\begin{aligned} & \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \\ & \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\end{aligned}\right\}\end{aligned}\) Where, $\rho \rightarrow$ Resistivity of wire $\mathrm{d} \rightarrow \text { density of wire }$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}}$ So, $\quad \mathrm{R}_{1} \propto \frac{l_{1}^{2}}{\mathrm{~m}_{1}}, \mathrm{R}_{2} \propto \frac{l_{2}^{2}}{\mathrm{~m}_{2}}$ and $\mathrm{R}_{3} \propto \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =\frac{l_{1}^{2}}{\mathrm{~m}_{1}}: \frac{l_{2}^{2}}{\mathrm{~m}_{2}}: \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}=25: 3: \frac{1}{5}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =125: 15: 1$
Karnataka CET-2019
Current Electricity
151866
The value of current (in Ampere) through the 3 $\Omega$ resistance in the circuit as shown is
1 $0.05 \mathrm{~A}$
2 $0.25 \mathrm{~A}$
3 $0.76 \mathrm{~A}$
4 $0.96 \mathrm{~A}$
Explanation:
D From figure, it is clear that $0.1 \Omega$ and $60 \Omega$ resistor are in parallel. So, $\frac{1}{\mathrm{R}} =\frac{1}{0.1}+\frac{1}{60}$ $=10+\frac{1}{60}$ $\frac{1}{\mathrm{R}} =\frac{601}{60}$ $\mathrm{R} =\frac{60}{601} \Omega$ Now, $R \Omega$ and $3 \Omega$ are in series combination. Hence, $\mathrm{R}^{\prime} =\mathrm{R}+3$ $=\frac{60}{601}+3=\frac{60+1803}{601}$ $\mathrm{R}^{\prime} =\frac{1863}{601}$ Current through $3 \Omega$ resistance, $i=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $=\frac{3}{\frac{1863}{601}}=\frac{1803}{1863}$ $\mathrm{i}=0.96 \mathrm{~A}$
AMU-2019
Current Electricity
151867
In the given circuit, find voltage across $12 \Omega$ resistance.
1 12 Volt
2 36 Volt
3 72 Volt
4 48 Volt
Explanation:
D After simplifying the circuit we get, $\mathrm{R}_{\text {eq }}=\frac{6 \times 12}{6+12}=4 \Omega$ $\mathrm{I}=12 \mathrm{~A}$ $\mathrm{~V}=\mathrm{IR}$ $\quad=12 \times 4$ $\mathrm{V}=48 \text { Volt. }$
JIPMER-2019
Current Electricity
151868
A circuit contain two resistance $R_{1}$ and $R_{2}$ in series. Find the ratio of input voltage to voltage of $\mathbf{R}_{\mathbf{2}}$.
B From figure, $\text { Input voltage, } \mathrm{V}=\mathrm{i}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)$ Voltage across $R_{2}, V_{R_{2}}=i R_{2}$ Required ratio $=\frac{\text { Input voltage }}{\text { voltage across } \mathrm{R}_{2}}$ $=\frac{V}{V_{R_{2}}}=\frac{i\left(R_{1}+R_{2}\right)}{i R_{2}}$ $=\frac{R_{1}+R_{2}}{R_{2}}$
JIPMER-2019
Current Electricity
151869
An electric cable has just one copper wire of radius $8 \mathrm{~mm}$ and resistance $10 \Omega$. This single copper wire cable is replaced by 4 different well insulated copper wires each of radius 6 $\mathrm{mm}$, then total resistance of the cable is
1 $4.44 \Omega$
2 $3.33 \Omega$
3 $2.22 \Omega$
4 $5.55 \Omega$
Explanation:
A Given, $\mathrm{r}_{1}=8 \mathrm{~mm}, \mathrm{r}_{2}=6 \mathrm{~mm}, \mathrm{R}_{1}=10 \Omega$ As we know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ Where, $\mathrm{R}=$ Resistance of wire $l=\text { length of wire }$ $\mathrm{A}=$ cross section of wire $\because \quad \mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\frac{10}{\mathrm{R}_{2}}=\frac{6^{2}}{8^{2}}=\frac{36}{64}$ $\frac{10}{\mathrm{R}_{2}}=\frac{36}{64}$ $\mathrm{R}_{2}=\frac{640}{36}$ $\mathrm{R}_{2}=\frac{160}{9}$ Equivalent resistance of 4 wire each of resistance $R_{2}$ connected in parallel is $\mathrm{R}^{\prime}$ $\mathrm{R}^{\prime}=\frac{\mathrm{R}_{2}}{4}=\frac{160}{9 \times 4}=4.44 \Omega$
151864
Masses of three wires of copper are in the ratio $1: 3: 5$ and their lengths are in the ratio $5: 3$ : 1. The ratio of their electrical resistance are :
1 $1: 3: 5$
2 $5: 3: 1$
3 $1: 15: 125$
4 $125: 15: 1$
Explanation:
D Given, Ratio of mass of wire $\mathrm{m}_{1}: \mathrm{m}_{2}: \mathrm{m}_{3}=1: 3: 5$ Ratio of length of wire $l_{1}: l_{2}: l_{3}=5: 3: 1$ We know that - $\text { Resistance }(\mathrm{R})=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}} \times \frac{l}{l}$ $\mathrm{R}=\rho \frac{l^{2}}{\mathrm{~V}}$ ${[\because \mathrm{A} \times l=\mathrm{V}]}$ $\mathrm{R}=\rho \frac{l^{2}}{\frac{\mathrm{m}}{\mathrm{d}}}$ \(\begin{aligned} \mathrm{R}=\rho \frac{l^2}{\mathrm{~V}} & {[\because \mathrm{A} \times l=\mathrm{V}] } \\ \mathrm{R}=\rho \frac{l^2}{\frac{\mathrm{m}}{\mathrm{d}}} & \left\{\begin{aligned} & \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \\ & \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\end{aligned}\right\}\end{aligned}\) Where, $\rho \rightarrow$ Resistivity of wire $\mathrm{d} \rightarrow \text { density of wire }$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}}$ So, $\quad \mathrm{R}_{1} \propto \frac{l_{1}^{2}}{\mathrm{~m}_{1}}, \mathrm{R}_{2} \propto \frac{l_{2}^{2}}{\mathrm{~m}_{2}}$ and $\mathrm{R}_{3} \propto \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =\frac{l_{1}^{2}}{\mathrm{~m}_{1}}: \frac{l_{2}^{2}}{\mathrm{~m}_{2}}: \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}=25: 3: \frac{1}{5}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =125: 15: 1$
Karnataka CET-2019
Current Electricity
151866
The value of current (in Ampere) through the 3 $\Omega$ resistance in the circuit as shown is
1 $0.05 \mathrm{~A}$
2 $0.25 \mathrm{~A}$
3 $0.76 \mathrm{~A}$
4 $0.96 \mathrm{~A}$
Explanation:
D From figure, it is clear that $0.1 \Omega$ and $60 \Omega$ resistor are in parallel. So, $\frac{1}{\mathrm{R}} =\frac{1}{0.1}+\frac{1}{60}$ $=10+\frac{1}{60}$ $\frac{1}{\mathrm{R}} =\frac{601}{60}$ $\mathrm{R} =\frac{60}{601} \Omega$ Now, $R \Omega$ and $3 \Omega$ are in series combination. Hence, $\mathrm{R}^{\prime} =\mathrm{R}+3$ $=\frac{60}{601}+3=\frac{60+1803}{601}$ $\mathrm{R}^{\prime} =\frac{1863}{601}$ Current through $3 \Omega$ resistance, $i=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $=\frac{3}{\frac{1863}{601}}=\frac{1803}{1863}$ $\mathrm{i}=0.96 \mathrm{~A}$
AMU-2019
Current Electricity
151867
In the given circuit, find voltage across $12 \Omega$ resistance.
1 12 Volt
2 36 Volt
3 72 Volt
4 48 Volt
Explanation:
D After simplifying the circuit we get, $\mathrm{R}_{\text {eq }}=\frac{6 \times 12}{6+12}=4 \Omega$ $\mathrm{I}=12 \mathrm{~A}$ $\mathrm{~V}=\mathrm{IR}$ $\quad=12 \times 4$ $\mathrm{V}=48 \text { Volt. }$
JIPMER-2019
Current Electricity
151868
A circuit contain two resistance $R_{1}$ and $R_{2}$ in series. Find the ratio of input voltage to voltage of $\mathbf{R}_{\mathbf{2}}$.
B From figure, $\text { Input voltage, } \mathrm{V}=\mathrm{i}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)$ Voltage across $R_{2}, V_{R_{2}}=i R_{2}$ Required ratio $=\frac{\text { Input voltage }}{\text { voltage across } \mathrm{R}_{2}}$ $=\frac{V}{V_{R_{2}}}=\frac{i\left(R_{1}+R_{2}\right)}{i R_{2}}$ $=\frac{R_{1}+R_{2}}{R_{2}}$
JIPMER-2019
Current Electricity
151869
An electric cable has just one copper wire of radius $8 \mathrm{~mm}$ and resistance $10 \Omega$. This single copper wire cable is replaced by 4 different well insulated copper wires each of radius 6 $\mathrm{mm}$, then total resistance of the cable is
1 $4.44 \Omega$
2 $3.33 \Omega$
3 $2.22 \Omega$
4 $5.55 \Omega$
Explanation:
A Given, $\mathrm{r}_{1}=8 \mathrm{~mm}, \mathrm{r}_{2}=6 \mathrm{~mm}, \mathrm{R}_{1}=10 \Omega$ As we know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ Where, $\mathrm{R}=$ Resistance of wire $l=\text { length of wire }$ $\mathrm{A}=$ cross section of wire $\because \quad \mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\frac{10}{\mathrm{R}_{2}}=\frac{6^{2}}{8^{2}}=\frac{36}{64}$ $\frac{10}{\mathrm{R}_{2}}=\frac{36}{64}$ $\mathrm{R}_{2}=\frac{640}{36}$ $\mathrm{R}_{2}=\frac{160}{9}$ Equivalent resistance of 4 wire each of resistance $R_{2}$ connected in parallel is $\mathrm{R}^{\prime}$ $\mathrm{R}^{\prime}=\frac{\mathrm{R}_{2}}{4}=\frac{160}{9 \times 4}=4.44 \Omega$
151864
Masses of three wires of copper are in the ratio $1: 3: 5$ and their lengths are in the ratio $5: 3$ : 1. The ratio of their electrical resistance are :
1 $1: 3: 5$
2 $5: 3: 1$
3 $1: 15: 125$
4 $125: 15: 1$
Explanation:
D Given, Ratio of mass of wire $\mathrm{m}_{1}: \mathrm{m}_{2}: \mathrm{m}_{3}=1: 3: 5$ Ratio of length of wire $l_{1}: l_{2}: l_{3}=5: 3: 1$ We know that - $\text { Resistance }(\mathrm{R})=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l}{\mathrm{~A}} \times \frac{l}{l}$ $\mathrm{R}=\rho \frac{l^{2}}{\mathrm{~V}}$ ${[\because \mathrm{A} \times l=\mathrm{V}]}$ $\mathrm{R}=\rho \frac{l^{2}}{\frac{\mathrm{m}}{\mathrm{d}}}$ \(\begin{aligned} \mathrm{R}=\rho \frac{l^2}{\mathrm{~V}} & {[\because \mathrm{A} \times l=\mathrm{V}] } \\ \mathrm{R}=\rho \frac{l^2}{\frac{\mathrm{m}}{\mathrm{d}}} & \left\{\begin{aligned} & \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \\ & \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\end{aligned}\right\}\end{aligned}\) Where, $\rho \rightarrow$ Resistivity of wire $\mathrm{d} \rightarrow \text { density of wire }$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}}$ So, $\quad \mathrm{R}_{1} \propto \frac{l_{1}^{2}}{\mathrm{~m}_{1}}, \mathrm{R}_{2} \propto \frac{l_{2}^{2}}{\mathrm{~m}_{2}}$ and $\mathrm{R}_{3} \propto \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =\frac{l_{1}^{2}}{\mathrm{~m}_{1}}: \frac{l_{2}^{2}}{\mathrm{~m}_{2}}: \frac{l_{3}^{2}}{\mathrm{~m}_{3}}$ $=\frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5}=25: 3: \frac{1}{5}$ $\mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3} =125: 15: 1$
Karnataka CET-2019
Current Electricity
151866
The value of current (in Ampere) through the 3 $\Omega$ resistance in the circuit as shown is
1 $0.05 \mathrm{~A}$
2 $0.25 \mathrm{~A}$
3 $0.76 \mathrm{~A}$
4 $0.96 \mathrm{~A}$
Explanation:
D From figure, it is clear that $0.1 \Omega$ and $60 \Omega$ resistor are in parallel. So, $\frac{1}{\mathrm{R}} =\frac{1}{0.1}+\frac{1}{60}$ $=10+\frac{1}{60}$ $\frac{1}{\mathrm{R}} =\frac{601}{60}$ $\mathrm{R} =\frac{60}{601} \Omega$ Now, $R \Omega$ and $3 \Omega$ are in series combination. Hence, $\mathrm{R}^{\prime} =\mathrm{R}+3$ $=\frac{60}{601}+3=\frac{60+1803}{601}$ $\mathrm{R}^{\prime} =\frac{1863}{601}$ Current through $3 \Omega$ resistance, $i=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $=\frac{3}{\frac{1863}{601}}=\frac{1803}{1863}$ $\mathrm{i}=0.96 \mathrm{~A}$
AMU-2019
Current Electricity
151867
In the given circuit, find voltage across $12 \Omega$ resistance.
1 12 Volt
2 36 Volt
3 72 Volt
4 48 Volt
Explanation:
D After simplifying the circuit we get, $\mathrm{R}_{\text {eq }}=\frac{6 \times 12}{6+12}=4 \Omega$ $\mathrm{I}=12 \mathrm{~A}$ $\mathrm{~V}=\mathrm{IR}$ $\quad=12 \times 4$ $\mathrm{V}=48 \text { Volt. }$
JIPMER-2019
Current Electricity
151868
A circuit contain two resistance $R_{1}$ and $R_{2}$ in series. Find the ratio of input voltage to voltage of $\mathbf{R}_{\mathbf{2}}$.
B From figure, $\text { Input voltage, } \mathrm{V}=\mathrm{i}\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)$ Voltage across $R_{2}, V_{R_{2}}=i R_{2}$ Required ratio $=\frac{\text { Input voltage }}{\text { voltage across } \mathrm{R}_{2}}$ $=\frac{V}{V_{R_{2}}}=\frac{i\left(R_{1}+R_{2}\right)}{i R_{2}}$ $=\frac{R_{1}+R_{2}}{R_{2}}$
JIPMER-2019
Current Electricity
151869
An electric cable has just one copper wire of radius $8 \mathrm{~mm}$ and resistance $10 \Omega$. This single copper wire cable is replaced by 4 different well insulated copper wires each of radius 6 $\mathrm{mm}$, then total resistance of the cable is
1 $4.44 \Omega$
2 $3.33 \Omega$
3 $2.22 \Omega$
4 $5.55 \Omega$
Explanation:
A Given, $\mathrm{r}_{1}=8 \mathrm{~mm}, \mathrm{r}_{2}=6 \mathrm{~mm}, \mathrm{R}_{1}=10 \Omega$ As we know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ Where, $\mathrm{R}=$ Resistance of wire $l=\text { length of wire }$ $\mathrm{A}=$ cross section of wire $\because \quad \mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\frac{10}{\mathrm{R}_{2}}=\frac{6^{2}}{8^{2}}=\frac{36}{64}$ $\frac{10}{\mathrm{R}_{2}}=\frac{36}{64}$ $\mathrm{R}_{2}=\frac{640}{36}$ $\mathrm{R}_{2}=\frac{160}{9}$ Equivalent resistance of 4 wire each of resistance $R_{2}$ connected in parallel is $\mathrm{R}^{\prime}$ $\mathrm{R}^{\prime}=\frac{\mathrm{R}_{2}}{4}=\frac{160}{9 \times 4}=4.44 \Omega$
