151719
A battery of $6 \mathrm{~V}$ is connected to the circuit as shown below. The current I drawn from the battery is:
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $\frac{6}{11} \mathrm{~A}$
4 $\frac{4}{3} \mathrm{~A}$
Explanation:
A Balance Wheat Stone Bridge in circuit so there in no current in $5 \Omega$ resistor so it can be removed from the circuit. $\mathrm{R}_{\mathrm{eq}}=\frac{6 \times 12}{6+12}+2$ $=\frac{72}{18}+2$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega$ $\therefore$ Current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{6}{6}=1 \mathrm{~A}$
JEE Main-26.07.2022
Current Electricity
151720
Two wires $A$ and $B$ of same material having length $L_{A}, L_{B}$ and radii $R_{A}, R_{B}$ and drift velocity $V_{A}, V_{B}$ respectively carries same current. If $L_{A}=L_{B}$ and $R_{A}=\mathbf{2} R_{B}$ then the value of $\left(\frac{V_{A}}{V_{B}}\right)$ is
1 0.25
2 0.5
3 2.0
4 1.0
Explanation:
A Given, Length of wire, $A=L_{A}$ Length of wire, $B=L_{B}$ Drift velocity of wire, $A=V_{A}$ Drift velocity of wire, $B=V_{B}$ And $\quad \mathrm{L}_{\mathrm{A}}=\mathrm{L}_{\mathrm{B}}, \quad \mathrm{R}_{\mathrm{A}}=2 \mathrm{R}_{\mathrm{B}}$ We know that, Current $(I)=$ nev $_{d} A$ So, $\quad \mathrm{I}_{1}=\mathrm{neV}_{\mathrm{A}} \pi \mathrm{R}_{\mathrm{A}}^{2}$ $\mathrm{I}_{2}=\mathrm{neV}_{\mathrm{B}} \pi \mathrm{R}_{\mathrm{B}}^{2}$ $\because \quad \mathrm{I}_{1}=\mathrm{I}_{2} \quad$ (given) Then, $\quad \mathrm{V}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^{2}=\mathrm{V}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\right)^{2}$ $\frac{V_{A}}{V_{B}}=\left(\frac{R_{B}}{2 R_{B}}\right)^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{2 \mathrm{R}_{\mathrm{B}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=0.25$
AP EAMCET-07.07.2022
Current Electricity
151721
The quantities that don't change when a resistor connected to a battery is heated due to the current are
1 $\mathrm{B}$ and $\mathrm{C}$
2 D
3 $\mathrm{A}$
4 A and D
Explanation:
B When the resistor is heated then drift speed decreases, resistivity and resistance increases but number of electrons remain same. Electrical resistivity is the reciprocal of electrical conductivity. It is the measure of the ability of a material to opposite the flow of current. Metals are good conductors of electricity. They have low resistivity. $\text { Resistivity }(\rho)=\frac{E}{J}$ Where, $\mathrm{E}=$ magnitude of the electric field $\left(\mathrm{Vm}^{-1}\right)$ $\mathrm{J}=$ Magnitude of current density $\left(\mathrm{Am}^{-2}\right)$
AP EAMCET-05.07.2022
Current Electricity
151722
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ ac mains supply. The peak current through the bulb will be:
1 $0.707 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $1.414 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A Given that, Power consume, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=200 \mathrm{~V}$ We know that, Resistance $\mathrm{R}=\mathrm{V}^{2} / \mathrm{P}={\frac{(200)^{2}}{100}}^{2}=400 \Omega$ $\mathrm{I}_{\mathrm{rms}}=\mathrm{V}_{\mathrm{rms}} / \mathrm{R}=\frac{200}{400}=0.5 \mathrm{~A}$ Peak current through the bulb $=0.5 \times \sqrt{2}=0.707 \mathrm{~A}$
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Current Electricity
151719
A battery of $6 \mathrm{~V}$ is connected to the circuit as shown below. The current I drawn from the battery is:
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $\frac{6}{11} \mathrm{~A}$
4 $\frac{4}{3} \mathrm{~A}$
Explanation:
A Balance Wheat Stone Bridge in circuit so there in no current in $5 \Omega$ resistor so it can be removed from the circuit. $\mathrm{R}_{\mathrm{eq}}=\frac{6 \times 12}{6+12}+2$ $=\frac{72}{18}+2$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega$ $\therefore$ Current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{6}{6}=1 \mathrm{~A}$
JEE Main-26.07.2022
Current Electricity
151720
Two wires $A$ and $B$ of same material having length $L_{A}, L_{B}$ and radii $R_{A}, R_{B}$ and drift velocity $V_{A}, V_{B}$ respectively carries same current. If $L_{A}=L_{B}$ and $R_{A}=\mathbf{2} R_{B}$ then the value of $\left(\frac{V_{A}}{V_{B}}\right)$ is
1 0.25
2 0.5
3 2.0
4 1.0
Explanation:
A Given, Length of wire, $A=L_{A}$ Length of wire, $B=L_{B}$ Drift velocity of wire, $A=V_{A}$ Drift velocity of wire, $B=V_{B}$ And $\quad \mathrm{L}_{\mathrm{A}}=\mathrm{L}_{\mathrm{B}}, \quad \mathrm{R}_{\mathrm{A}}=2 \mathrm{R}_{\mathrm{B}}$ We know that, Current $(I)=$ nev $_{d} A$ So, $\quad \mathrm{I}_{1}=\mathrm{neV}_{\mathrm{A}} \pi \mathrm{R}_{\mathrm{A}}^{2}$ $\mathrm{I}_{2}=\mathrm{neV}_{\mathrm{B}} \pi \mathrm{R}_{\mathrm{B}}^{2}$ $\because \quad \mathrm{I}_{1}=\mathrm{I}_{2} \quad$ (given) Then, $\quad \mathrm{V}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^{2}=\mathrm{V}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\right)^{2}$ $\frac{V_{A}}{V_{B}}=\left(\frac{R_{B}}{2 R_{B}}\right)^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{2 \mathrm{R}_{\mathrm{B}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=0.25$
AP EAMCET-07.07.2022
Current Electricity
151721
The quantities that don't change when a resistor connected to a battery is heated due to the current are
1 $\mathrm{B}$ and $\mathrm{C}$
2 D
3 $\mathrm{A}$
4 A and D
Explanation:
B When the resistor is heated then drift speed decreases, resistivity and resistance increases but number of electrons remain same. Electrical resistivity is the reciprocal of electrical conductivity. It is the measure of the ability of a material to opposite the flow of current. Metals are good conductors of electricity. They have low resistivity. $\text { Resistivity }(\rho)=\frac{E}{J}$ Where, $\mathrm{E}=$ magnitude of the electric field $\left(\mathrm{Vm}^{-1}\right)$ $\mathrm{J}=$ Magnitude of current density $\left(\mathrm{Am}^{-2}\right)$
AP EAMCET-05.07.2022
Current Electricity
151722
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ ac mains supply. The peak current through the bulb will be:
1 $0.707 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $1.414 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A Given that, Power consume, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=200 \mathrm{~V}$ We know that, Resistance $\mathrm{R}=\mathrm{V}^{2} / \mathrm{P}={\frac{(200)^{2}}{100}}^{2}=400 \Omega$ $\mathrm{I}_{\mathrm{rms}}=\mathrm{V}_{\mathrm{rms}} / \mathrm{R}=\frac{200}{400}=0.5 \mathrm{~A}$ Peak current through the bulb $=0.5 \times \sqrt{2}=0.707 \mathrm{~A}$
151719
A battery of $6 \mathrm{~V}$ is connected to the circuit as shown below. The current I drawn from the battery is:
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $\frac{6}{11} \mathrm{~A}$
4 $\frac{4}{3} \mathrm{~A}$
Explanation:
A Balance Wheat Stone Bridge in circuit so there in no current in $5 \Omega$ resistor so it can be removed from the circuit. $\mathrm{R}_{\mathrm{eq}}=\frac{6 \times 12}{6+12}+2$ $=\frac{72}{18}+2$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega$ $\therefore$ Current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{6}{6}=1 \mathrm{~A}$
JEE Main-26.07.2022
Current Electricity
151720
Two wires $A$ and $B$ of same material having length $L_{A}, L_{B}$ and radii $R_{A}, R_{B}$ and drift velocity $V_{A}, V_{B}$ respectively carries same current. If $L_{A}=L_{B}$ and $R_{A}=\mathbf{2} R_{B}$ then the value of $\left(\frac{V_{A}}{V_{B}}\right)$ is
1 0.25
2 0.5
3 2.0
4 1.0
Explanation:
A Given, Length of wire, $A=L_{A}$ Length of wire, $B=L_{B}$ Drift velocity of wire, $A=V_{A}$ Drift velocity of wire, $B=V_{B}$ And $\quad \mathrm{L}_{\mathrm{A}}=\mathrm{L}_{\mathrm{B}}, \quad \mathrm{R}_{\mathrm{A}}=2 \mathrm{R}_{\mathrm{B}}$ We know that, Current $(I)=$ nev $_{d} A$ So, $\quad \mathrm{I}_{1}=\mathrm{neV}_{\mathrm{A}} \pi \mathrm{R}_{\mathrm{A}}^{2}$ $\mathrm{I}_{2}=\mathrm{neV}_{\mathrm{B}} \pi \mathrm{R}_{\mathrm{B}}^{2}$ $\because \quad \mathrm{I}_{1}=\mathrm{I}_{2} \quad$ (given) Then, $\quad \mathrm{V}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^{2}=\mathrm{V}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\right)^{2}$ $\frac{V_{A}}{V_{B}}=\left(\frac{R_{B}}{2 R_{B}}\right)^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{2 \mathrm{R}_{\mathrm{B}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=0.25$
AP EAMCET-07.07.2022
Current Electricity
151721
The quantities that don't change when a resistor connected to a battery is heated due to the current are
1 $\mathrm{B}$ and $\mathrm{C}$
2 D
3 $\mathrm{A}$
4 A and D
Explanation:
B When the resistor is heated then drift speed decreases, resistivity and resistance increases but number of electrons remain same. Electrical resistivity is the reciprocal of electrical conductivity. It is the measure of the ability of a material to opposite the flow of current. Metals are good conductors of electricity. They have low resistivity. $\text { Resistivity }(\rho)=\frac{E}{J}$ Where, $\mathrm{E}=$ magnitude of the electric field $\left(\mathrm{Vm}^{-1}\right)$ $\mathrm{J}=$ Magnitude of current density $\left(\mathrm{Am}^{-2}\right)$
AP EAMCET-05.07.2022
Current Electricity
151722
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ ac mains supply. The peak current through the bulb will be:
1 $0.707 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $1.414 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A Given that, Power consume, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=200 \mathrm{~V}$ We know that, Resistance $\mathrm{R}=\mathrm{V}^{2} / \mathrm{P}={\frac{(200)^{2}}{100}}^{2}=400 \Omega$ $\mathrm{I}_{\mathrm{rms}}=\mathrm{V}_{\mathrm{rms}} / \mathrm{R}=\frac{200}{400}=0.5 \mathrm{~A}$ Peak current through the bulb $=0.5 \times \sqrt{2}=0.707 \mathrm{~A}$
151719
A battery of $6 \mathrm{~V}$ is connected to the circuit as shown below. The current I drawn from the battery is:
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $\frac{6}{11} \mathrm{~A}$
4 $\frac{4}{3} \mathrm{~A}$
Explanation:
A Balance Wheat Stone Bridge in circuit so there in no current in $5 \Omega$ resistor so it can be removed from the circuit. $\mathrm{R}_{\mathrm{eq}}=\frac{6 \times 12}{6+12}+2$ $=\frac{72}{18}+2$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega$ $\therefore$ Current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{6}{6}=1 \mathrm{~A}$
JEE Main-26.07.2022
Current Electricity
151720
Two wires $A$ and $B$ of same material having length $L_{A}, L_{B}$ and radii $R_{A}, R_{B}$ and drift velocity $V_{A}, V_{B}$ respectively carries same current. If $L_{A}=L_{B}$ and $R_{A}=\mathbf{2} R_{B}$ then the value of $\left(\frac{V_{A}}{V_{B}}\right)$ is
1 0.25
2 0.5
3 2.0
4 1.0
Explanation:
A Given, Length of wire, $A=L_{A}$ Length of wire, $B=L_{B}$ Drift velocity of wire, $A=V_{A}$ Drift velocity of wire, $B=V_{B}$ And $\quad \mathrm{L}_{\mathrm{A}}=\mathrm{L}_{\mathrm{B}}, \quad \mathrm{R}_{\mathrm{A}}=2 \mathrm{R}_{\mathrm{B}}$ We know that, Current $(I)=$ nev $_{d} A$ So, $\quad \mathrm{I}_{1}=\mathrm{neV}_{\mathrm{A}} \pi \mathrm{R}_{\mathrm{A}}^{2}$ $\mathrm{I}_{2}=\mathrm{neV}_{\mathrm{B}} \pi \mathrm{R}_{\mathrm{B}}^{2}$ $\because \quad \mathrm{I}_{1}=\mathrm{I}_{2} \quad$ (given) Then, $\quad \mathrm{V}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^{2}=\mathrm{V}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}\right)^{2}$ $\frac{V_{A}}{V_{B}}=\left(\frac{R_{B}}{2 R_{B}}\right)^{2}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{B}}}{2 \mathrm{R}_{\mathrm{B}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=0.25$
AP EAMCET-07.07.2022
Current Electricity
151721
The quantities that don't change when a resistor connected to a battery is heated due to the current are
1 $\mathrm{B}$ and $\mathrm{C}$
2 D
3 $\mathrm{A}$
4 A and D
Explanation:
B When the resistor is heated then drift speed decreases, resistivity and resistance increases but number of electrons remain same. Electrical resistivity is the reciprocal of electrical conductivity. It is the measure of the ability of a material to opposite the flow of current. Metals are good conductors of electricity. They have low resistivity. $\text { Resistivity }(\rho)=\frac{E}{J}$ Where, $\mathrm{E}=$ magnitude of the electric field $\left(\mathrm{Vm}^{-1}\right)$ $\mathrm{J}=$ Magnitude of current density $\left(\mathrm{Am}^{-2}\right)$
AP EAMCET-05.07.2022
Current Electricity
151722
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ ac mains supply. The peak current through the bulb will be:
1 $0.707 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $1.414 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A Given that, Power consume, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=200 \mathrm{~V}$ We know that, Resistance $\mathrm{R}=\mathrm{V}^{2} / \mathrm{P}={\frac{(200)^{2}}{100}}^{2}=400 \Omega$ $\mathrm{I}_{\mathrm{rms}}=\mathrm{V}_{\mathrm{rms}} / \mathrm{R}=\frac{200}{400}=0.5 \mathrm{~A}$ Peak current through the bulb $=0.5 \times \sqrt{2}=0.707 \mathrm{~A}$
