152880
Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. The bridge is balanced when $C=100 \Omega$. If $A$ and $B$ are interchanged, the bridge balances for $\mathrm{C}=$ $121 \Omega$. The value of $D$ is
1 $10 \Omega$
2 $100 \Omega$
3 $110 \Omega$
4 $120 \Omega$
Explanation:
C The balanced condition of Wheatstone bridge, $\frac{A}{B}=\frac{C}{D}$ $\frac{A}{B}=\frac{100}{D}$ $D=\frac{100 B}{A}$ Interchanging the resistance $\mathrm{B}$ and $\mathrm{A}$, we are getting a new value of bridge balance condition for $\mathrm{C}$ $\mathrm{C}=121 \Omega$ The new balance condition $\frac{B}{A}=\frac{C}{D}$ $\frac{B}{A}=\frac{121}{D}$ Form eq ${ }^{\mathrm{n}}$. (i) $\mathrm{D}=\frac{100 \mathrm{~B}}{\mathrm{~A}}$ $\mathrm{D}=100 \times \frac{121}{\mathrm{D}}$ Rearranging the terms we get. $\mathrm{D}^{2} =100 \times 121$ $\mathrm{D} =\sqrt{100 \times 121}$ $=10 \times 11$ $=110 \Omega$ $\mathrm{D} =110 \Omega$
AP EAMCET (20.04.2019) Shift-II
Current Electricity
152881
The length of a potentiometer wire is $\ell$. A cell of emf $E$ is balanced at a length $\ell / 3$ from the positive end of the wire. If the length of the wire is increased by $\ell / 2$. Then at what distance will be the same cell give a balanced point.
1 $2 \ell / 3$
2 $\ell / 2$
3 $\ell / 6$
4 $4 \ell / 3$
Explanation:
B For cell of emf E balancing length $=\frac{l}{3}=\frac{1}{3} \times l($ length of wire $)$ Increased length of wire $=l+\frac{l}{2}=\frac{3 l}{2}$ Since, cell is same, therefore the new balancing length $=\frac{1}{3} \times(\text { new increasing length })$ $=\frac{1}{3} \times \frac{3}{2} l=\frac{l}{2}$
AP EAMCET (22.04.2019) Shift-I
Current Electricity
152883
The resistance of $1 \mathrm{~A}$ ammeter is $0.018 \Omega$. To convert it into 10 A ammeter, the shunt resistance required will be
1 $0.18 \Omega$
2 $0.0018 \Omega$
3 $0.002 \Omega$
4 $0.12 \Omega$
Explanation:
C Given, $I_{g}=1 A, \quad R=0.018 \Omega$ $I=10 A, S=?$ We know that, $\left(I-I_{g}\right) S =I_{g} R$ $(10-1) S=1 \times 0.018$ $S =\frac{0.018}{9}$ $S =0.002 \Omega$
CG PET 2019
Current Electricity
152884
In the circuits shown below, the readings of voltmeters and the ammeters will be
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
2 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}=\mathrm{i}_{2}$
3 $\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
4 $V_{2}>V_{2}$ and $i_{1}=i_{2}$
Explanation:
B Form above figure, $\mathrm{R}_{\mathrm{v}}=\infty \Omega$ $\mathrm{R}_{\mathrm{A}}=0 \Omega$ For (I)- $\frac{1}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{1}{10}+\frac{1}{\mathrm{R}_{\mathrm{V}}}=\frac{1}{10}+\frac{1}{\infty}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{10}{10}=1 \mathrm{~A}$ For (II)- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=0+\frac{1}{10}+\frac{1}{(10+\infty)}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{2}=\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{eq}_{2}}}=\frac{10}{10}=1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1}=\mathrm{i}_{2}$ $\mathrm{V}_{1}=1 \times 10=10 \mathrm{~V}$ $\mathrm{~V}_{1}=\mathrm{V}_{2} \text { and } \mathrm{i}_{1}=\mathrm{i}_{2}$
152880
Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. The bridge is balanced when $C=100 \Omega$. If $A$ and $B$ are interchanged, the bridge balances for $\mathrm{C}=$ $121 \Omega$. The value of $D$ is
1 $10 \Omega$
2 $100 \Omega$
3 $110 \Omega$
4 $120 \Omega$
Explanation:
C The balanced condition of Wheatstone bridge, $\frac{A}{B}=\frac{C}{D}$ $\frac{A}{B}=\frac{100}{D}$ $D=\frac{100 B}{A}$ Interchanging the resistance $\mathrm{B}$ and $\mathrm{A}$, we are getting a new value of bridge balance condition for $\mathrm{C}$ $\mathrm{C}=121 \Omega$ The new balance condition $\frac{B}{A}=\frac{C}{D}$ $\frac{B}{A}=\frac{121}{D}$ Form eq ${ }^{\mathrm{n}}$. (i) $\mathrm{D}=\frac{100 \mathrm{~B}}{\mathrm{~A}}$ $\mathrm{D}=100 \times \frac{121}{\mathrm{D}}$ Rearranging the terms we get. $\mathrm{D}^{2} =100 \times 121$ $\mathrm{D} =\sqrt{100 \times 121}$ $=10 \times 11$ $=110 \Omega$ $\mathrm{D} =110 \Omega$
AP EAMCET (20.04.2019) Shift-II
Current Electricity
152881
The length of a potentiometer wire is $\ell$. A cell of emf $E$ is balanced at a length $\ell / 3$ from the positive end of the wire. If the length of the wire is increased by $\ell / 2$. Then at what distance will be the same cell give a balanced point.
1 $2 \ell / 3$
2 $\ell / 2$
3 $\ell / 6$
4 $4 \ell / 3$
Explanation:
B For cell of emf E balancing length $=\frac{l}{3}=\frac{1}{3} \times l($ length of wire $)$ Increased length of wire $=l+\frac{l}{2}=\frac{3 l}{2}$ Since, cell is same, therefore the new balancing length $=\frac{1}{3} \times(\text { new increasing length })$ $=\frac{1}{3} \times \frac{3}{2} l=\frac{l}{2}$
AP EAMCET (22.04.2019) Shift-I
Current Electricity
152883
The resistance of $1 \mathrm{~A}$ ammeter is $0.018 \Omega$. To convert it into 10 A ammeter, the shunt resistance required will be
1 $0.18 \Omega$
2 $0.0018 \Omega$
3 $0.002 \Omega$
4 $0.12 \Omega$
Explanation:
C Given, $I_{g}=1 A, \quad R=0.018 \Omega$ $I=10 A, S=?$ We know that, $\left(I-I_{g}\right) S =I_{g} R$ $(10-1) S=1 \times 0.018$ $S =\frac{0.018}{9}$ $S =0.002 \Omega$
CG PET 2019
Current Electricity
152884
In the circuits shown below, the readings of voltmeters and the ammeters will be
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
2 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}=\mathrm{i}_{2}$
3 $\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
4 $V_{2}>V_{2}$ and $i_{1}=i_{2}$
Explanation:
B Form above figure, $\mathrm{R}_{\mathrm{v}}=\infty \Omega$ $\mathrm{R}_{\mathrm{A}}=0 \Omega$ For (I)- $\frac{1}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{1}{10}+\frac{1}{\mathrm{R}_{\mathrm{V}}}=\frac{1}{10}+\frac{1}{\infty}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{10}{10}=1 \mathrm{~A}$ For (II)- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=0+\frac{1}{10}+\frac{1}{(10+\infty)}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{2}=\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{eq}_{2}}}=\frac{10}{10}=1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1}=\mathrm{i}_{2}$ $\mathrm{V}_{1}=1 \times 10=10 \mathrm{~V}$ $\mathrm{~V}_{1}=\mathrm{V}_{2} \text { and } \mathrm{i}_{1}=\mathrm{i}_{2}$
152880
Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. The bridge is balanced when $C=100 \Omega$. If $A$ and $B$ are interchanged, the bridge balances for $\mathrm{C}=$ $121 \Omega$. The value of $D$ is
1 $10 \Omega$
2 $100 \Omega$
3 $110 \Omega$
4 $120 \Omega$
Explanation:
C The balanced condition of Wheatstone bridge, $\frac{A}{B}=\frac{C}{D}$ $\frac{A}{B}=\frac{100}{D}$ $D=\frac{100 B}{A}$ Interchanging the resistance $\mathrm{B}$ and $\mathrm{A}$, we are getting a new value of bridge balance condition for $\mathrm{C}$ $\mathrm{C}=121 \Omega$ The new balance condition $\frac{B}{A}=\frac{C}{D}$ $\frac{B}{A}=\frac{121}{D}$ Form eq ${ }^{\mathrm{n}}$. (i) $\mathrm{D}=\frac{100 \mathrm{~B}}{\mathrm{~A}}$ $\mathrm{D}=100 \times \frac{121}{\mathrm{D}}$ Rearranging the terms we get. $\mathrm{D}^{2} =100 \times 121$ $\mathrm{D} =\sqrt{100 \times 121}$ $=10 \times 11$ $=110 \Omega$ $\mathrm{D} =110 \Omega$
AP EAMCET (20.04.2019) Shift-II
Current Electricity
152881
The length of a potentiometer wire is $\ell$. A cell of emf $E$ is balanced at a length $\ell / 3$ from the positive end of the wire. If the length of the wire is increased by $\ell / 2$. Then at what distance will be the same cell give a balanced point.
1 $2 \ell / 3$
2 $\ell / 2$
3 $\ell / 6$
4 $4 \ell / 3$
Explanation:
B For cell of emf E balancing length $=\frac{l}{3}=\frac{1}{3} \times l($ length of wire $)$ Increased length of wire $=l+\frac{l}{2}=\frac{3 l}{2}$ Since, cell is same, therefore the new balancing length $=\frac{1}{3} \times(\text { new increasing length })$ $=\frac{1}{3} \times \frac{3}{2} l=\frac{l}{2}$
AP EAMCET (22.04.2019) Shift-I
Current Electricity
152883
The resistance of $1 \mathrm{~A}$ ammeter is $0.018 \Omega$. To convert it into 10 A ammeter, the shunt resistance required will be
1 $0.18 \Omega$
2 $0.0018 \Omega$
3 $0.002 \Omega$
4 $0.12 \Omega$
Explanation:
C Given, $I_{g}=1 A, \quad R=0.018 \Omega$ $I=10 A, S=?$ We know that, $\left(I-I_{g}\right) S =I_{g} R$ $(10-1) S=1 \times 0.018$ $S =\frac{0.018}{9}$ $S =0.002 \Omega$
CG PET 2019
Current Electricity
152884
In the circuits shown below, the readings of voltmeters and the ammeters will be
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
2 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}=\mathrm{i}_{2}$
3 $\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
4 $V_{2}>V_{2}$ and $i_{1}=i_{2}$
Explanation:
B Form above figure, $\mathrm{R}_{\mathrm{v}}=\infty \Omega$ $\mathrm{R}_{\mathrm{A}}=0 \Omega$ For (I)- $\frac{1}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{1}{10}+\frac{1}{\mathrm{R}_{\mathrm{V}}}=\frac{1}{10}+\frac{1}{\infty}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{10}{10}=1 \mathrm{~A}$ For (II)- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=0+\frac{1}{10}+\frac{1}{(10+\infty)}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{2}=\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{eq}_{2}}}=\frac{10}{10}=1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1}=\mathrm{i}_{2}$ $\mathrm{V}_{1}=1 \times 10=10 \mathrm{~V}$ $\mathrm{~V}_{1}=\mathrm{V}_{2} \text { and } \mathrm{i}_{1}=\mathrm{i}_{2}$
152880
Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. The bridge is balanced when $C=100 \Omega$. If $A$ and $B$ are interchanged, the bridge balances for $\mathrm{C}=$ $121 \Omega$. The value of $D$ is
1 $10 \Omega$
2 $100 \Omega$
3 $110 \Omega$
4 $120 \Omega$
Explanation:
C The balanced condition of Wheatstone bridge, $\frac{A}{B}=\frac{C}{D}$ $\frac{A}{B}=\frac{100}{D}$ $D=\frac{100 B}{A}$ Interchanging the resistance $\mathrm{B}$ and $\mathrm{A}$, we are getting a new value of bridge balance condition for $\mathrm{C}$ $\mathrm{C}=121 \Omega$ The new balance condition $\frac{B}{A}=\frac{C}{D}$ $\frac{B}{A}=\frac{121}{D}$ Form eq ${ }^{\mathrm{n}}$. (i) $\mathrm{D}=\frac{100 \mathrm{~B}}{\mathrm{~A}}$ $\mathrm{D}=100 \times \frac{121}{\mathrm{D}}$ Rearranging the terms we get. $\mathrm{D}^{2} =100 \times 121$ $\mathrm{D} =\sqrt{100 \times 121}$ $=10 \times 11$ $=110 \Omega$ $\mathrm{D} =110 \Omega$
AP EAMCET (20.04.2019) Shift-II
Current Electricity
152881
The length of a potentiometer wire is $\ell$. A cell of emf $E$ is balanced at a length $\ell / 3$ from the positive end of the wire. If the length of the wire is increased by $\ell / 2$. Then at what distance will be the same cell give a balanced point.
1 $2 \ell / 3$
2 $\ell / 2$
3 $\ell / 6$
4 $4 \ell / 3$
Explanation:
B For cell of emf E balancing length $=\frac{l}{3}=\frac{1}{3} \times l($ length of wire $)$ Increased length of wire $=l+\frac{l}{2}=\frac{3 l}{2}$ Since, cell is same, therefore the new balancing length $=\frac{1}{3} \times(\text { new increasing length })$ $=\frac{1}{3} \times \frac{3}{2} l=\frac{l}{2}$
AP EAMCET (22.04.2019) Shift-I
Current Electricity
152883
The resistance of $1 \mathrm{~A}$ ammeter is $0.018 \Omega$. To convert it into 10 A ammeter, the shunt resistance required will be
1 $0.18 \Omega$
2 $0.0018 \Omega$
3 $0.002 \Omega$
4 $0.12 \Omega$
Explanation:
C Given, $I_{g}=1 A, \quad R=0.018 \Omega$ $I=10 A, S=?$ We know that, $\left(I-I_{g}\right) S =I_{g} R$ $(10-1) S=1 \times 0.018$ $S =\frac{0.018}{9}$ $S =0.002 \Omega$
CG PET 2019
Current Electricity
152884
In the circuits shown below, the readings of voltmeters and the ammeters will be
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
2 $\mathrm{V}_{1}=\mathrm{V}_{2}$ and $\mathrm{i}_{1}=\mathrm{i}_{2}$
3 $\mathrm{V}_{2}>\mathrm{V}_{1}$ and $\mathrm{i}_{1}>\mathrm{i}_{2}$
4 $V_{2}>V_{2}$ and $i_{1}=i_{2}$
Explanation:
B Form above figure, $\mathrm{R}_{\mathrm{v}}=\infty \Omega$ $\mathrm{R}_{\mathrm{A}}=0 \Omega$ For (I)- $\frac{1}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{1}{10}+\frac{1}{\mathrm{R}_{\mathrm{V}}}=\frac{1}{10}+\frac{1}{\infty}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}_{1}}}=\frac{10}{10}=1 \mathrm{~A}$ For (II)- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=0+\frac{1}{10}+\frac{1}{(10+\infty)}$ $\mathrm{R}_{\mathrm{eq}}=10 \Omega$ $\mathrm{i}_{2}=\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{eq}_{2}}}=\frac{10}{10}=1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1}=\mathrm{i}_{2}$ $\mathrm{V}_{1}=1 \times 10=10 \mathrm{~V}$ $\mathrm{~V}_{1}=\mathrm{V}_{2} \text { and } \mathrm{i}_{1}=\mathrm{i}_{2}$
