152194
A current of $3 \mathrm{~A}$ flows through the following circuit in anticlockwise direction as well as in clockwise direction. The value of $E$ respectively is
1 $6 \mathrm{~V}, 10 \mathrm{~V}$
2 $1 \mathrm{~V}, 19 \mathrm{~V}$
3 $4 \mathrm{~V}, 8 \mathrm{~V}$
4 $3 \mathrm{~V}, 7 \mathrm{~V}$
Explanation:
B Case-I: when $3 \mathrm{~A}$ current is flowing clockwise Apply KVL in the circuit, $\begin{aligned} & E-(2 \times 3)-10-1 \times 3=0 \\ & E-6-10-3=0 \\ & E=19 V \end{aligned}$ Case-II: when current flow in anti clockwise direction $\begin{aligned} & E+(2 \times 3)-10+(3 \times 1)=0 \\ & E+6-10+3=0 \\ & E=1 V \end{aligned}$ The value of $\mathrm{E}$ is $1 \mathrm{~V}, 19 \mathrm{~V}$
MHT-CET 2020
Current Electricity
152196
In the following electrical network, the value of ' $I$ ' is
1 $2.3 \mathrm{~A}$
2 $3.4 \mathrm{~A}$
3 $1.8 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
B Applying $\mathrm{KCl}$ at junction $\mathrm{O}$, Applying $\mathrm{KCl}$ at junction $\mathrm{P}$
MHT-CET 2020
Current Electricity
152197
The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
1 $2.4 \mathrm{~A}$
2 $3.6 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.2 \mathrm{~A}$
Explanation:
A Simplify this circuit The above arrangement balanced according to Wheatstone bridge. Equivalent resistance, $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{5}+\frac{1}{5} \\ & \mathrm{R}=\frac{5}{2} \Omega \end{aligned}$ Current drawn from battery $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152198
The value of current (I) in the given current distribution is
1 $0.6 \mathrm{~A}$
2 $0.4 \mathrm{~A}$
3 $0.7 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Explanation:
C Applying Kirchhoff's current law $\mathrm{I}+0.2+0.4-0.8-0.5=0$ $\mathrm{I}+0.6-1.3=0$ $\mathrm{I}=1.3-0.6$ $\mathrm{I}=0.7 \mathrm{~A}$
152194
A current of $3 \mathrm{~A}$ flows through the following circuit in anticlockwise direction as well as in clockwise direction. The value of $E$ respectively is
1 $6 \mathrm{~V}, 10 \mathrm{~V}$
2 $1 \mathrm{~V}, 19 \mathrm{~V}$
3 $4 \mathrm{~V}, 8 \mathrm{~V}$
4 $3 \mathrm{~V}, 7 \mathrm{~V}$
Explanation:
B Case-I: when $3 \mathrm{~A}$ current is flowing clockwise Apply KVL in the circuit, $\begin{aligned} & E-(2 \times 3)-10-1 \times 3=0 \\ & E-6-10-3=0 \\ & E=19 V \end{aligned}$ Case-II: when current flow in anti clockwise direction $\begin{aligned} & E+(2 \times 3)-10+(3 \times 1)=0 \\ & E+6-10+3=0 \\ & E=1 V \end{aligned}$ The value of $\mathrm{E}$ is $1 \mathrm{~V}, 19 \mathrm{~V}$
MHT-CET 2020
Current Electricity
152196
In the following electrical network, the value of ' $I$ ' is
1 $2.3 \mathrm{~A}$
2 $3.4 \mathrm{~A}$
3 $1.8 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
B Applying $\mathrm{KCl}$ at junction $\mathrm{O}$, Applying $\mathrm{KCl}$ at junction $\mathrm{P}$
MHT-CET 2020
Current Electricity
152197
The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
1 $2.4 \mathrm{~A}$
2 $3.6 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.2 \mathrm{~A}$
Explanation:
A Simplify this circuit The above arrangement balanced according to Wheatstone bridge. Equivalent resistance, $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{5}+\frac{1}{5} \\ & \mathrm{R}=\frac{5}{2} \Omega \end{aligned}$ Current drawn from battery $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152198
The value of current (I) in the given current distribution is
1 $0.6 \mathrm{~A}$
2 $0.4 \mathrm{~A}$
3 $0.7 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Explanation:
C Applying Kirchhoff's current law $\mathrm{I}+0.2+0.4-0.8-0.5=0$ $\mathrm{I}+0.6-1.3=0$ $\mathrm{I}=1.3-0.6$ $\mathrm{I}=0.7 \mathrm{~A}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152194
A current of $3 \mathrm{~A}$ flows through the following circuit in anticlockwise direction as well as in clockwise direction. The value of $E$ respectively is
1 $6 \mathrm{~V}, 10 \mathrm{~V}$
2 $1 \mathrm{~V}, 19 \mathrm{~V}$
3 $4 \mathrm{~V}, 8 \mathrm{~V}$
4 $3 \mathrm{~V}, 7 \mathrm{~V}$
Explanation:
B Case-I: when $3 \mathrm{~A}$ current is flowing clockwise Apply KVL in the circuit, $\begin{aligned} & E-(2 \times 3)-10-1 \times 3=0 \\ & E-6-10-3=0 \\ & E=19 V \end{aligned}$ Case-II: when current flow in anti clockwise direction $\begin{aligned} & E+(2 \times 3)-10+(3 \times 1)=0 \\ & E+6-10+3=0 \\ & E=1 V \end{aligned}$ The value of $\mathrm{E}$ is $1 \mathrm{~V}, 19 \mathrm{~V}$
MHT-CET 2020
Current Electricity
152196
In the following electrical network, the value of ' $I$ ' is
1 $2.3 \mathrm{~A}$
2 $3.4 \mathrm{~A}$
3 $1.8 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
B Applying $\mathrm{KCl}$ at junction $\mathrm{O}$, Applying $\mathrm{KCl}$ at junction $\mathrm{P}$
MHT-CET 2020
Current Electricity
152197
The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
1 $2.4 \mathrm{~A}$
2 $3.6 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.2 \mathrm{~A}$
Explanation:
A Simplify this circuit The above arrangement balanced according to Wheatstone bridge. Equivalent resistance, $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{5}+\frac{1}{5} \\ & \mathrm{R}=\frac{5}{2} \Omega \end{aligned}$ Current drawn from battery $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152198
The value of current (I) in the given current distribution is
1 $0.6 \mathrm{~A}$
2 $0.4 \mathrm{~A}$
3 $0.7 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Explanation:
C Applying Kirchhoff's current law $\mathrm{I}+0.2+0.4-0.8-0.5=0$ $\mathrm{I}+0.6-1.3=0$ $\mathrm{I}=1.3-0.6$ $\mathrm{I}=0.7 \mathrm{~A}$
152194
A current of $3 \mathrm{~A}$ flows through the following circuit in anticlockwise direction as well as in clockwise direction. The value of $E$ respectively is
1 $6 \mathrm{~V}, 10 \mathrm{~V}$
2 $1 \mathrm{~V}, 19 \mathrm{~V}$
3 $4 \mathrm{~V}, 8 \mathrm{~V}$
4 $3 \mathrm{~V}, 7 \mathrm{~V}$
Explanation:
B Case-I: when $3 \mathrm{~A}$ current is flowing clockwise Apply KVL in the circuit, $\begin{aligned} & E-(2 \times 3)-10-1 \times 3=0 \\ & E-6-10-3=0 \\ & E=19 V \end{aligned}$ Case-II: when current flow in anti clockwise direction $\begin{aligned} & E+(2 \times 3)-10+(3 \times 1)=0 \\ & E+6-10+3=0 \\ & E=1 V \end{aligned}$ The value of $\mathrm{E}$ is $1 \mathrm{~V}, 19 \mathrm{~V}$
MHT-CET 2020
Current Electricity
152196
In the following electrical network, the value of ' $I$ ' is
1 $2.3 \mathrm{~A}$
2 $3.4 \mathrm{~A}$
3 $1.8 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
B Applying $\mathrm{KCl}$ at junction $\mathrm{O}$, Applying $\mathrm{KCl}$ at junction $\mathrm{P}$
MHT-CET 2020
Current Electricity
152197
The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
1 $2.4 \mathrm{~A}$
2 $3.6 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.2 \mathrm{~A}$
Explanation:
A Simplify this circuit The above arrangement balanced according to Wheatstone bridge. Equivalent resistance, $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{5}+\frac{1}{5} \\ & \mathrm{R}=\frac{5}{2} \Omega \end{aligned}$ Current drawn from battery $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152198
The value of current (I) in the given current distribution is
1 $0.6 \mathrm{~A}$
2 $0.4 \mathrm{~A}$
3 $0.7 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Explanation:
C Applying Kirchhoff's current law $\mathrm{I}+0.2+0.4-0.8-0.5=0$ $\mathrm{I}+0.6-1.3=0$ $\mathrm{I}=1.3-0.6$ $\mathrm{I}=0.7 \mathrm{~A}$
