151886
Find $V_{P}-V_{Q}$ in the circuit shown in figure.
1 $6.68 \mathrm{~V}$
2 $8 \mathrm{~V}$
3 $4.65 \mathrm{~V}$
4 $7 \mathrm{~V}$
Explanation:
C $10 \Omega$ and $8 \Omega$ are parallel between P \& Q $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $10 \Omega, 18 \Omega$ and $25 \Omega$ are parallel between $Q \& \mathrm{R}$ $\frac{1}{\mathrm{R}_{\mathrm{QR}}}=\frac{1}{10}+\frac{1}{18}+\frac{1}{25} =\frac{45+25+18}{450}$ $=\frac{88}{450}=\frac{44}{225}$ $\mathrm{R}_{\mathrm{QR}}=\frac{225}{44} \Omega$ Now, $\quad \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{PQ}}+\mathrm{R}_{\mathrm{QR}}=\frac{40}{9}+\frac{225}{44}$ $=9.56 \Omega$ $\therefore \quad \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{10}{9.56}=1.05 \mathrm{~A}$ Then, $\quad \mathrm{V}_{\mathrm{PQ}}=\mathrm{i} \times \mathrm{R}_{\mathrm{PQ}}=1.05 \times \frac{40}{9}$ $=4.65 \mathrm{~V}$
JIPMER-2018
Current Electricity
151887
$A$ and $B$ are two points on a uniform ring of resistance $19 \Omega$. The angle $\angle \mathrm{AOB}=45^{\circ}$. The equivalent resistance between $A$ and $B$ is -
1 $8.02 \Omega$
2 $20.8 \Omega$
3 $3.8 \Omega$
4 $2.082 \Omega$
Explanation:
D Resistance of $360^{\circ}(\mathrm{R})=19 \Omega$ Resistance of $1^{\circ}=\frac{19}{360^{\circ}}$ Resistance of $45^{\circ}=\frac{19}{360^{\circ}} \times 45^{\circ}$ Then, $\mathrm{R}_{2}=\mathrm{R}-\mathrm{R}_{1}=19-2.375=16.625$ $\mathrm{R}_{\text {са }}=\frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{R}_{\text {еq }}=\frac{2.375 \times 16.625}{2.375+16.625}=\frac{39.484}{19}=2.0781 \square 2.082 \Omega$
AP EAMCET-25.04.2018
Current Electricity
151888
Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$, the ratio of their electrical resistances $\left(\mathbf{R}_{A} / \mathbf{R}_{B}\right)$ is
1 0.04
2 25
3 625
4 225
Explanation:
C Given, $\mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}, l_{\mathrm{A}}=25 l_{\mathrm{B}}$ As we know that, Resistance, $\quad \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R} =\frac{\rho l \cdot l}{\mathrm{~V}}$ $\mathrm{R} =\frac{\rho l^{2} \cdot \mathrm{d}}{\mathrm{m}} \quad\left[\therefore \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\right]$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}} \ldots . . \text { (i) }$ $\mathrm{R}_{\mathrm{A}} \propto \frac{l_{\mathrm{A}}^{2}}{\mathrm{~m}_{\mathrm{A}}} \ldots . \text { (ii) }$ On dividing equation (i) and (ii), we get $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(l_{\mathrm{A}}\right)^{2} \times \mathrm{m}_{\mathrm{B}}}{\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{(25)^{2}\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}{\left(l_{\mathrm{B}}\right)^{2} \mathrm{~m}_{\mathrm{A}}}=625$
TS EAMCET 02.05.2018
Current Electricity
151890
What will be current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is made on?
1 Zero
2 $100 \mathrm{~mA}$
3 $10 \mathrm{~mA}$
4 $1 \mathrm{~mA}$
Explanation:
C The circuit is in steady state, So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 \Omega$ and $400 \Omega$ is same. Now, equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=200+400=600 \Omega$ $\therefore$ Current through $200 \Omega$ So, current in circuit, $I =\frac{V}{R_{\text {eq }}}$ $I =\frac{6}{600}$ $I =0.01 \mathrm{~A}=10 \mathrm{~mA}$
WB JEE 2018
Current Electricity
151891
Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?
1 $210 \Omega$
2 $240 \Omega$
3 $300 \Omega$
4 $250 \Omega$
Explanation:
D For maximum equivalent resistance across the diagonal of the square, the given resistors connected as, Resistance of ACB arm, $\mathrm{R}_{1}=300+200=500 \Omega$ Resistance of ADB arm, $\mathrm{R}_{2}=400+100 \Omega=500 \Omega$ The equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{500}+\frac{1}{500}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{500}$ $\mathrm{R}_{\mathrm{eq}}=250 \Omega$
151886
Find $V_{P}-V_{Q}$ in the circuit shown in figure.
1 $6.68 \mathrm{~V}$
2 $8 \mathrm{~V}$
3 $4.65 \mathrm{~V}$
4 $7 \mathrm{~V}$
Explanation:
C $10 \Omega$ and $8 \Omega$ are parallel between P \& Q $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $10 \Omega, 18 \Omega$ and $25 \Omega$ are parallel between $Q \& \mathrm{R}$ $\frac{1}{\mathrm{R}_{\mathrm{QR}}}=\frac{1}{10}+\frac{1}{18}+\frac{1}{25} =\frac{45+25+18}{450}$ $=\frac{88}{450}=\frac{44}{225}$ $\mathrm{R}_{\mathrm{QR}}=\frac{225}{44} \Omega$ Now, $\quad \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{PQ}}+\mathrm{R}_{\mathrm{QR}}=\frac{40}{9}+\frac{225}{44}$ $=9.56 \Omega$ $\therefore \quad \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{10}{9.56}=1.05 \mathrm{~A}$ Then, $\quad \mathrm{V}_{\mathrm{PQ}}=\mathrm{i} \times \mathrm{R}_{\mathrm{PQ}}=1.05 \times \frac{40}{9}$ $=4.65 \mathrm{~V}$
JIPMER-2018
Current Electricity
151887
$A$ and $B$ are two points on a uniform ring of resistance $19 \Omega$. The angle $\angle \mathrm{AOB}=45^{\circ}$. The equivalent resistance between $A$ and $B$ is -
1 $8.02 \Omega$
2 $20.8 \Omega$
3 $3.8 \Omega$
4 $2.082 \Omega$
Explanation:
D Resistance of $360^{\circ}(\mathrm{R})=19 \Omega$ Resistance of $1^{\circ}=\frac{19}{360^{\circ}}$ Resistance of $45^{\circ}=\frac{19}{360^{\circ}} \times 45^{\circ}$ Then, $\mathrm{R}_{2}=\mathrm{R}-\mathrm{R}_{1}=19-2.375=16.625$ $\mathrm{R}_{\text {са }}=\frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{R}_{\text {еq }}=\frac{2.375 \times 16.625}{2.375+16.625}=\frac{39.484}{19}=2.0781 \square 2.082 \Omega$
AP EAMCET-25.04.2018
Current Electricity
151888
Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$, the ratio of their electrical resistances $\left(\mathbf{R}_{A} / \mathbf{R}_{B}\right)$ is
1 0.04
2 25
3 625
4 225
Explanation:
C Given, $\mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}, l_{\mathrm{A}}=25 l_{\mathrm{B}}$ As we know that, Resistance, $\quad \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R} =\frac{\rho l \cdot l}{\mathrm{~V}}$ $\mathrm{R} =\frac{\rho l^{2} \cdot \mathrm{d}}{\mathrm{m}} \quad\left[\therefore \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\right]$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}} \ldots . . \text { (i) }$ $\mathrm{R}_{\mathrm{A}} \propto \frac{l_{\mathrm{A}}^{2}}{\mathrm{~m}_{\mathrm{A}}} \ldots . \text { (ii) }$ On dividing equation (i) and (ii), we get $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(l_{\mathrm{A}}\right)^{2} \times \mathrm{m}_{\mathrm{B}}}{\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{(25)^{2}\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}{\left(l_{\mathrm{B}}\right)^{2} \mathrm{~m}_{\mathrm{A}}}=625$
TS EAMCET 02.05.2018
Current Electricity
151890
What will be current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is made on?
1 Zero
2 $100 \mathrm{~mA}$
3 $10 \mathrm{~mA}$
4 $1 \mathrm{~mA}$
Explanation:
C The circuit is in steady state, So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 \Omega$ and $400 \Omega$ is same. Now, equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=200+400=600 \Omega$ $\therefore$ Current through $200 \Omega$ So, current in circuit, $I =\frac{V}{R_{\text {eq }}}$ $I =\frac{6}{600}$ $I =0.01 \mathrm{~A}=10 \mathrm{~mA}$
WB JEE 2018
Current Electricity
151891
Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?
1 $210 \Omega$
2 $240 \Omega$
3 $300 \Omega$
4 $250 \Omega$
Explanation:
D For maximum equivalent resistance across the diagonal of the square, the given resistors connected as, Resistance of ACB arm, $\mathrm{R}_{1}=300+200=500 \Omega$ Resistance of ADB arm, $\mathrm{R}_{2}=400+100 \Omega=500 \Omega$ The equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{500}+\frac{1}{500}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{500}$ $\mathrm{R}_{\mathrm{eq}}=250 \Omega$
151886
Find $V_{P}-V_{Q}$ in the circuit shown in figure.
1 $6.68 \mathrm{~V}$
2 $8 \mathrm{~V}$
3 $4.65 \mathrm{~V}$
4 $7 \mathrm{~V}$
Explanation:
C $10 \Omega$ and $8 \Omega$ are parallel between P \& Q $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $10 \Omega, 18 \Omega$ and $25 \Omega$ are parallel between $Q \& \mathrm{R}$ $\frac{1}{\mathrm{R}_{\mathrm{QR}}}=\frac{1}{10}+\frac{1}{18}+\frac{1}{25} =\frac{45+25+18}{450}$ $=\frac{88}{450}=\frac{44}{225}$ $\mathrm{R}_{\mathrm{QR}}=\frac{225}{44} \Omega$ Now, $\quad \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{PQ}}+\mathrm{R}_{\mathrm{QR}}=\frac{40}{9}+\frac{225}{44}$ $=9.56 \Omega$ $\therefore \quad \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{10}{9.56}=1.05 \mathrm{~A}$ Then, $\quad \mathrm{V}_{\mathrm{PQ}}=\mathrm{i} \times \mathrm{R}_{\mathrm{PQ}}=1.05 \times \frac{40}{9}$ $=4.65 \mathrm{~V}$
JIPMER-2018
Current Electricity
151887
$A$ and $B$ are two points on a uniform ring of resistance $19 \Omega$. The angle $\angle \mathrm{AOB}=45^{\circ}$. The equivalent resistance between $A$ and $B$ is -
1 $8.02 \Omega$
2 $20.8 \Omega$
3 $3.8 \Omega$
4 $2.082 \Omega$
Explanation:
D Resistance of $360^{\circ}(\mathrm{R})=19 \Omega$ Resistance of $1^{\circ}=\frac{19}{360^{\circ}}$ Resistance of $45^{\circ}=\frac{19}{360^{\circ}} \times 45^{\circ}$ Then, $\mathrm{R}_{2}=\mathrm{R}-\mathrm{R}_{1}=19-2.375=16.625$ $\mathrm{R}_{\text {са }}=\frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{R}_{\text {еq }}=\frac{2.375 \times 16.625}{2.375+16.625}=\frac{39.484}{19}=2.0781 \square 2.082 \Omega$
AP EAMCET-25.04.2018
Current Electricity
151888
Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$, the ratio of their electrical resistances $\left(\mathbf{R}_{A} / \mathbf{R}_{B}\right)$ is
1 0.04
2 25
3 625
4 225
Explanation:
C Given, $\mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}, l_{\mathrm{A}}=25 l_{\mathrm{B}}$ As we know that, Resistance, $\quad \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R} =\frac{\rho l \cdot l}{\mathrm{~V}}$ $\mathrm{R} =\frac{\rho l^{2} \cdot \mathrm{d}}{\mathrm{m}} \quad\left[\therefore \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\right]$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}} \ldots . . \text { (i) }$ $\mathrm{R}_{\mathrm{A}} \propto \frac{l_{\mathrm{A}}^{2}}{\mathrm{~m}_{\mathrm{A}}} \ldots . \text { (ii) }$ On dividing equation (i) and (ii), we get $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(l_{\mathrm{A}}\right)^{2} \times \mathrm{m}_{\mathrm{B}}}{\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{(25)^{2}\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}{\left(l_{\mathrm{B}}\right)^{2} \mathrm{~m}_{\mathrm{A}}}=625$
TS EAMCET 02.05.2018
Current Electricity
151890
What will be current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is made on?
1 Zero
2 $100 \mathrm{~mA}$
3 $10 \mathrm{~mA}$
4 $1 \mathrm{~mA}$
Explanation:
C The circuit is in steady state, So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 \Omega$ and $400 \Omega$ is same. Now, equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=200+400=600 \Omega$ $\therefore$ Current through $200 \Omega$ So, current in circuit, $I =\frac{V}{R_{\text {eq }}}$ $I =\frac{6}{600}$ $I =0.01 \mathrm{~A}=10 \mathrm{~mA}$
WB JEE 2018
Current Electricity
151891
Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?
1 $210 \Omega$
2 $240 \Omega$
3 $300 \Omega$
4 $250 \Omega$
Explanation:
D For maximum equivalent resistance across the diagonal of the square, the given resistors connected as, Resistance of ACB arm, $\mathrm{R}_{1}=300+200=500 \Omega$ Resistance of ADB arm, $\mathrm{R}_{2}=400+100 \Omega=500 \Omega$ The equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{500}+\frac{1}{500}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{500}$ $\mathrm{R}_{\mathrm{eq}}=250 \Omega$
151886
Find $V_{P}-V_{Q}$ in the circuit shown in figure.
1 $6.68 \mathrm{~V}$
2 $8 \mathrm{~V}$
3 $4.65 \mathrm{~V}$
4 $7 \mathrm{~V}$
Explanation:
C $10 \Omega$ and $8 \Omega$ are parallel between P \& Q $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $10 \Omega, 18 \Omega$ and $25 \Omega$ are parallel between $Q \& \mathrm{R}$ $\frac{1}{\mathrm{R}_{\mathrm{QR}}}=\frac{1}{10}+\frac{1}{18}+\frac{1}{25} =\frac{45+25+18}{450}$ $=\frac{88}{450}=\frac{44}{225}$ $\mathrm{R}_{\mathrm{QR}}=\frac{225}{44} \Omega$ Now, $\quad \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{PQ}}+\mathrm{R}_{\mathrm{QR}}=\frac{40}{9}+\frac{225}{44}$ $=9.56 \Omega$ $\therefore \quad \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{10}{9.56}=1.05 \mathrm{~A}$ Then, $\quad \mathrm{V}_{\mathrm{PQ}}=\mathrm{i} \times \mathrm{R}_{\mathrm{PQ}}=1.05 \times \frac{40}{9}$ $=4.65 \mathrm{~V}$
JIPMER-2018
Current Electricity
151887
$A$ and $B$ are two points on a uniform ring of resistance $19 \Omega$. The angle $\angle \mathrm{AOB}=45^{\circ}$. The equivalent resistance between $A$ and $B$ is -
1 $8.02 \Omega$
2 $20.8 \Omega$
3 $3.8 \Omega$
4 $2.082 \Omega$
Explanation:
D Resistance of $360^{\circ}(\mathrm{R})=19 \Omega$ Resistance of $1^{\circ}=\frac{19}{360^{\circ}}$ Resistance of $45^{\circ}=\frac{19}{360^{\circ}} \times 45^{\circ}$ Then, $\mathrm{R}_{2}=\mathrm{R}-\mathrm{R}_{1}=19-2.375=16.625$ $\mathrm{R}_{\text {са }}=\frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{R}_{\text {еq }}=\frac{2.375 \times 16.625}{2.375+16.625}=\frac{39.484}{19}=2.0781 \square 2.082 \Omega$
AP EAMCET-25.04.2018
Current Electricity
151888
Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$, the ratio of their electrical resistances $\left(\mathbf{R}_{A} / \mathbf{R}_{B}\right)$ is
1 0.04
2 25
3 625
4 225
Explanation:
C Given, $\mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}, l_{\mathrm{A}}=25 l_{\mathrm{B}}$ As we know that, Resistance, $\quad \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R} =\frac{\rho l \cdot l}{\mathrm{~V}}$ $\mathrm{R} =\frac{\rho l^{2} \cdot \mathrm{d}}{\mathrm{m}} \quad\left[\therefore \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\right]$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}} \ldots . . \text { (i) }$ $\mathrm{R}_{\mathrm{A}} \propto \frac{l_{\mathrm{A}}^{2}}{\mathrm{~m}_{\mathrm{A}}} \ldots . \text { (ii) }$ On dividing equation (i) and (ii), we get $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(l_{\mathrm{A}}\right)^{2} \times \mathrm{m}_{\mathrm{B}}}{\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{(25)^{2}\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}{\left(l_{\mathrm{B}}\right)^{2} \mathrm{~m}_{\mathrm{A}}}=625$
TS EAMCET 02.05.2018
Current Electricity
151890
What will be current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is made on?
1 Zero
2 $100 \mathrm{~mA}$
3 $10 \mathrm{~mA}$
4 $1 \mathrm{~mA}$
Explanation:
C The circuit is in steady state, So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 \Omega$ and $400 \Omega$ is same. Now, equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=200+400=600 \Omega$ $\therefore$ Current through $200 \Omega$ So, current in circuit, $I =\frac{V}{R_{\text {eq }}}$ $I =\frac{6}{600}$ $I =0.01 \mathrm{~A}=10 \mathrm{~mA}$
WB JEE 2018
Current Electricity
151891
Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?
1 $210 \Omega$
2 $240 \Omega$
3 $300 \Omega$
4 $250 \Omega$
Explanation:
D For maximum equivalent resistance across the diagonal of the square, the given resistors connected as, Resistance of ACB arm, $\mathrm{R}_{1}=300+200=500 \Omega$ Resistance of ADB arm, $\mathrm{R}_{2}=400+100 \Omega=500 \Omega$ The equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{500}+\frac{1}{500}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{500}$ $\mathrm{R}_{\mathrm{eq}}=250 \Omega$
151886
Find $V_{P}-V_{Q}$ in the circuit shown in figure.
1 $6.68 \mathrm{~V}$
2 $8 \mathrm{~V}$
3 $4.65 \mathrm{~V}$
4 $7 \mathrm{~V}$
Explanation:
C $10 \Omega$ and $8 \Omega$ are parallel between P \& Q $\mathrm{R}_{\mathrm{PQ}}=\frac{10 \times 8}{10+8}=\frac{80}{18}=\frac{40}{9} \Omega$ $10 \Omega, 18 \Omega$ and $25 \Omega$ are parallel between $Q \& \mathrm{R}$ $\frac{1}{\mathrm{R}_{\mathrm{QR}}}=\frac{1}{10}+\frac{1}{18}+\frac{1}{25} =\frac{45+25+18}{450}$ $=\frac{88}{450}=\frac{44}{225}$ $\mathrm{R}_{\mathrm{QR}}=\frac{225}{44} \Omega$ Now, $\quad \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{PQ}}+\mathrm{R}_{\mathrm{QR}}=\frac{40}{9}+\frac{225}{44}$ $=9.56 \Omega$ $\therefore \quad \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{10}{9.56}=1.05 \mathrm{~A}$ Then, $\quad \mathrm{V}_{\mathrm{PQ}}=\mathrm{i} \times \mathrm{R}_{\mathrm{PQ}}=1.05 \times \frac{40}{9}$ $=4.65 \mathrm{~V}$
JIPMER-2018
Current Electricity
151887
$A$ and $B$ are two points on a uniform ring of resistance $19 \Omega$. The angle $\angle \mathrm{AOB}=45^{\circ}$. The equivalent resistance between $A$ and $B$ is -
1 $8.02 \Omega$
2 $20.8 \Omega$
3 $3.8 \Omega$
4 $2.082 \Omega$
Explanation:
D Resistance of $360^{\circ}(\mathrm{R})=19 \Omega$ Resistance of $1^{\circ}=\frac{19}{360^{\circ}}$ Resistance of $45^{\circ}=\frac{19}{360^{\circ}} \times 45^{\circ}$ Then, $\mathrm{R}_{2}=\mathrm{R}-\mathrm{R}_{1}=19-2.375=16.625$ $\mathrm{R}_{\text {са }}=\frac{\mathrm{R}_{1} \times \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{R}_{\text {еq }}=\frac{2.375 \times 16.625}{2.375+16.625}=\frac{39.484}{19}=2.0781 \square 2.082 \Omega$
AP EAMCET-25.04.2018
Current Electricity
151888
Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$, the ratio of their electrical resistances $\left(\mathbf{R}_{A} / \mathbf{R}_{B}\right)$ is
1 0.04
2 25
3 625
4 225
Explanation:
C Given, $\mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}, l_{\mathrm{A}}=25 l_{\mathrm{B}}$ As we know that, Resistance, $\quad \mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R} =\frac{\rho l \cdot l}{\mathrm{~V}}$ $\mathrm{R} =\frac{\rho l^{2} \cdot \mathrm{d}}{\mathrm{m}} \quad\left[\therefore \mathrm{V}=\frac{\mathrm{m}}{\mathrm{d}}\right]$ $\mathrm{R} \propto \frac{l^{2}}{\mathrm{~m}} \ldots . . \text { (i) }$ $\mathrm{R}_{\mathrm{A}} \propto \frac{l_{\mathrm{A}}^{2}}{\mathrm{~m}_{\mathrm{A}}} \ldots . \text { (ii) }$ On dividing equation (i) and (ii), we get $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(l_{\mathrm{A}}\right)^{2} \times \mathrm{m}_{\mathrm{B}}}{\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{(25)^{2}\left(l_{\mathrm{B}}\right)^{2} \times \mathrm{m}_{\mathrm{A}}}{\left(l_{\mathrm{B}}\right)^{2} \mathrm{~m}_{\mathrm{A}}}=625$
TS EAMCET 02.05.2018
Current Electricity
151890
What will be current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is made on?
1 Zero
2 $100 \mathrm{~mA}$
3 $10 \mathrm{~mA}$
4 $1 \mathrm{~mA}$
Explanation:
C The circuit is in steady state, So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 \Omega$ and $400 \Omega$ is same. Now, equivalent resistance, $\mathrm{R}_{\mathrm{eq}}=200+400=600 \Omega$ $\therefore$ Current through $200 \Omega$ So, current in circuit, $I =\frac{V}{R_{\text {eq }}}$ $I =\frac{6}{600}$ $I =0.01 \mathrm{~A}=10 \mathrm{~mA}$
WB JEE 2018
Current Electricity
151891
Four resistors, $100 \Omega, 200 \Omega, 300 \Omega$ and $400 \Omega$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?
1 $210 \Omega$
2 $240 \Omega$
3 $300 \Omega$
4 $250 \Omega$
Explanation:
D For maximum equivalent resistance across the diagonal of the square, the given resistors connected as, Resistance of ACB arm, $\mathrm{R}_{1}=300+200=500 \Omega$ Resistance of ADB arm, $\mathrm{R}_{2}=400+100 \Omega=500 \Omega$ The equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{500}+\frac{1}{500}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2}{500}$ $\mathrm{R}_{\mathrm{eq}}=250 \Omega$
