NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152180
Conservation of charge and conservation of energy are respectively the basis of
1 Joule's law and Ampere's circuital law
2 Gauss' law and Ohm's law
3 Kirchhoff's junction rule and loop rule
4 Coulomb's inverse square law and Gauiss' law
5 Joule's law and Ohm's law
Explanation:
C Kirchhoff's first law -} It states that the sum of current entering to the junction is equal to sum of current leaving the junction. i.e. $\quad \sum \mathrm{I}=0$ The first law of Kirchhoff's is based on charge conservation. Now, Kirchhoff's second law or loop rule is based on conservation of energy, as the sum of potential drops in a closed circuit is zero. $\sum \mathrm{IR}=0$ We can also write - Now, we can say that energy is conserved becomes a circuit loop is a closed conducting path. So, no energy is lost. Hence, Kirchhoff's junction rule and loop rule are based on charge and energy conservation.
Kerala CEE 04.07.2022
Current Electricity
152181 For the given following circuit diagram, the dissipated of electrical power $150 \mathrm{~W}$, then find value of Resistance $R=$
1 $5 \Omega$
2 $8 \Omega$
3 $6 \Omega$
4 $3 \Omega$
Explanation:
C We know that, The equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=\frac{2 \times \mathrm{R}}{2+\mathrm{R}}$ Power dissipation $(P)=\frac{V^{2}}{R_{e q}}$ $\begin{aligned} 150 & =\frac{15 \times 15}{\mathrm{R}_{\text {eq }}} \\ \mathrm{R}_{\mathrm{eq}} & =\frac{15 \times 15}{150}=\frac{15}{10}=\frac{3}{2} \end{aligned}$ Therefore, $\frac{2 \times \mathrm{R}}{2+\mathrm{R}}=\frac{3}{2}$ $\begin{aligned} & 4 R=6+3 R \\ & R=6 \Omega \end{aligned}$
GUJCET 18.04.2022
Current Electricity
152182
If each of the resistance in the circuit is $10 \Omega$, the resistance between terminals $A$ and $B$ is
1 $10 \Omega$
2 $20 \Omega$
3 $30 \Omega$
4 $50 \Omega$
Explanation:
A Equivalent resistance between $\mathrm{APR}=10 \Omega+10 \Omega=20 \Omega$ Equivalent resistance between $\mathrm{AQR}=10 \Omega+10 \Omega$ $=20 \Omega$ The two resistances (i.e. $20 \Omega$ and $20 \Omega$ ) are in parallel. Hence equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ is given by $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10} \\ & \mathrm{R}=10 \Omega \end{aligned}$
Shift-I]
Current Electricity
152183
With reference to the given combination of resistances, the equivalent resistance $R_{A B}$ between points $A$ and $B$ is
1 $\frac{18}{5} \Omega$
2 $\frac{16}{3} \Omega$
3 $3 \Omega$
4 $5 \Omega$
Explanation:
A Given circuit, Circuit is balanced by wheat stone bridge so no current flow in centre. $\mathrm{R}_{\mathrm{AB}}=\frac{6 \times 9}{9+6}=\frac{54}{15}=\frac{18}{5} \Omega$
152180
Conservation of charge and conservation of energy are respectively the basis of
1 Joule's law and Ampere's circuital law
2 Gauss' law and Ohm's law
3 Kirchhoff's junction rule and loop rule
4 Coulomb's inverse square law and Gauiss' law
5 Joule's law and Ohm's law
Explanation:
C Kirchhoff's first law -} It states that the sum of current entering to the junction is equal to sum of current leaving the junction. i.e. $\quad \sum \mathrm{I}=0$ The first law of Kirchhoff's is based on charge conservation. Now, Kirchhoff's second law or loop rule is based on conservation of energy, as the sum of potential drops in a closed circuit is zero. $\sum \mathrm{IR}=0$ We can also write - Now, we can say that energy is conserved becomes a circuit loop is a closed conducting path. So, no energy is lost. Hence, Kirchhoff's junction rule and loop rule are based on charge and energy conservation.
Kerala CEE 04.07.2022
Current Electricity
152181 For the given following circuit diagram, the dissipated of electrical power $150 \mathrm{~W}$, then find value of Resistance $R=$
1 $5 \Omega$
2 $8 \Omega$
3 $6 \Omega$
4 $3 \Omega$
Explanation:
C We know that, The equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=\frac{2 \times \mathrm{R}}{2+\mathrm{R}}$ Power dissipation $(P)=\frac{V^{2}}{R_{e q}}$ $\begin{aligned} 150 & =\frac{15 \times 15}{\mathrm{R}_{\text {eq }}} \\ \mathrm{R}_{\mathrm{eq}} & =\frac{15 \times 15}{150}=\frac{15}{10}=\frac{3}{2} \end{aligned}$ Therefore, $\frac{2 \times \mathrm{R}}{2+\mathrm{R}}=\frac{3}{2}$ $\begin{aligned} & 4 R=6+3 R \\ & R=6 \Omega \end{aligned}$
GUJCET 18.04.2022
Current Electricity
152182
If each of the resistance in the circuit is $10 \Omega$, the resistance between terminals $A$ and $B$ is
1 $10 \Omega$
2 $20 \Omega$
3 $30 \Omega$
4 $50 \Omega$
Explanation:
A Equivalent resistance between $\mathrm{APR}=10 \Omega+10 \Omega=20 \Omega$ Equivalent resistance between $\mathrm{AQR}=10 \Omega+10 \Omega$ $=20 \Omega$ The two resistances (i.e. $20 \Omega$ and $20 \Omega$ ) are in parallel. Hence equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ is given by $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10} \\ & \mathrm{R}=10 \Omega \end{aligned}$
Shift-I]
Current Electricity
152183
With reference to the given combination of resistances, the equivalent resistance $R_{A B}$ between points $A$ and $B$ is
1 $\frac{18}{5} \Omega$
2 $\frac{16}{3} \Omega$
3 $3 \Omega$
4 $5 \Omega$
Explanation:
A Given circuit, Circuit is balanced by wheat stone bridge so no current flow in centre. $\mathrm{R}_{\mathrm{AB}}=\frac{6 \times 9}{9+6}=\frac{54}{15}=\frac{18}{5} \Omega$
152180
Conservation of charge and conservation of energy are respectively the basis of
1 Joule's law and Ampere's circuital law
2 Gauss' law and Ohm's law
3 Kirchhoff's junction rule and loop rule
4 Coulomb's inverse square law and Gauiss' law
5 Joule's law and Ohm's law
Explanation:
C Kirchhoff's first law -} It states that the sum of current entering to the junction is equal to sum of current leaving the junction. i.e. $\quad \sum \mathrm{I}=0$ The first law of Kirchhoff's is based on charge conservation. Now, Kirchhoff's second law or loop rule is based on conservation of energy, as the sum of potential drops in a closed circuit is zero. $\sum \mathrm{IR}=0$ We can also write - Now, we can say that energy is conserved becomes a circuit loop is a closed conducting path. So, no energy is lost. Hence, Kirchhoff's junction rule and loop rule are based on charge and energy conservation.
Kerala CEE 04.07.2022
Current Electricity
152181 For the given following circuit diagram, the dissipated of electrical power $150 \mathrm{~W}$, then find value of Resistance $R=$
1 $5 \Omega$
2 $8 \Omega$
3 $6 \Omega$
4 $3 \Omega$
Explanation:
C We know that, The equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=\frac{2 \times \mathrm{R}}{2+\mathrm{R}}$ Power dissipation $(P)=\frac{V^{2}}{R_{e q}}$ $\begin{aligned} 150 & =\frac{15 \times 15}{\mathrm{R}_{\text {eq }}} \\ \mathrm{R}_{\mathrm{eq}} & =\frac{15 \times 15}{150}=\frac{15}{10}=\frac{3}{2} \end{aligned}$ Therefore, $\frac{2 \times \mathrm{R}}{2+\mathrm{R}}=\frac{3}{2}$ $\begin{aligned} & 4 R=6+3 R \\ & R=6 \Omega \end{aligned}$
GUJCET 18.04.2022
Current Electricity
152182
If each of the resistance in the circuit is $10 \Omega$, the resistance between terminals $A$ and $B$ is
1 $10 \Omega$
2 $20 \Omega$
3 $30 \Omega$
4 $50 \Omega$
Explanation:
A Equivalent resistance between $\mathrm{APR}=10 \Omega+10 \Omega=20 \Omega$ Equivalent resistance between $\mathrm{AQR}=10 \Omega+10 \Omega$ $=20 \Omega$ The two resistances (i.e. $20 \Omega$ and $20 \Omega$ ) are in parallel. Hence equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ is given by $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10} \\ & \mathrm{R}=10 \Omega \end{aligned}$
Shift-I]
Current Electricity
152183
With reference to the given combination of resistances, the equivalent resistance $R_{A B}$ between points $A$ and $B$ is
1 $\frac{18}{5} \Omega$
2 $\frac{16}{3} \Omega$
3 $3 \Omega$
4 $5 \Omega$
Explanation:
A Given circuit, Circuit is balanced by wheat stone bridge so no current flow in centre. $\mathrm{R}_{\mathrm{AB}}=\frac{6 \times 9}{9+6}=\frac{54}{15}=\frac{18}{5} \Omega$
152180
Conservation of charge and conservation of energy are respectively the basis of
1 Joule's law and Ampere's circuital law
2 Gauss' law and Ohm's law
3 Kirchhoff's junction rule and loop rule
4 Coulomb's inverse square law and Gauiss' law
5 Joule's law and Ohm's law
Explanation:
C Kirchhoff's first law -} It states that the sum of current entering to the junction is equal to sum of current leaving the junction. i.e. $\quad \sum \mathrm{I}=0$ The first law of Kirchhoff's is based on charge conservation. Now, Kirchhoff's second law or loop rule is based on conservation of energy, as the sum of potential drops in a closed circuit is zero. $\sum \mathrm{IR}=0$ We can also write - Now, we can say that energy is conserved becomes a circuit loop is a closed conducting path. So, no energy is lost. Hence, Kirchhoff's junction rule and loop rule are based on charge and energy conservation.
Kerala CEE 04.07.2022
Current Electricity
152181 For the given following circuit diagram, the dissipated of electrical power $150 \mathrm{~W}$, then find value of Resistance $R=$
1 $5 \Omega$
2 $8 \Omega$
3 $6 \Omega$
4 $3 \Omega$
Explanation:
C We know that, The equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=\frac{2 \times \mathrm{R}}{2+\mathrm{R}}$ Power dissipation $(P)=\frac{V^{2}}{R_{e q}}$ $\begin{aligned} 150 & =\frac{15 \times 15}{\mathrm{R}_{\text {eq }}} \\ \mathrm{R}_{\mathrm{eq}} & =\frac{15 \times 15}{150}=\frac{15}{10}=\frac{3}{2} \end{aligned}$ Therefore, $\frac{2 \times \mathrm{R}}{2+\mathrm{R}}=\frac{3}{2}$ $\begin{aligned} & 4 R=6+3 R \\ & R=6 \Omega \end{aligned}$
GUJCET 18.04.2022
Current Electricity
152182
If each of the resistance in the circuit is $10 \Omega$, the resistance between terminals $A$ and $B$ is
1 $10 \Omega$
2 $20 \Omega$
3 $30 \Omega$
4 $50 \Omega$
Explanation:
A Equivalent resistance between $\mathrm{APR}=10 \Omega+10 \Omega=20 \Omega$ Equivalent resistance between $\mathrm{AQR}=10 \Omega+10 \Omega$ $=20 \Omega$ The two resistances (i.e. $20 \Omega$ and $20 \Omega$ ) are in parallel. Hence equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$ is given by $\begin{aligned} & \frac{1}{\mathrm{R}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10} \\ & \mathrm{R}=10 \Omega \end{aligned}$
Shift-I]
Current Electricity
152183
With reference to the given combination of resistances, the equivalent resistance $R_{A B}$ between points $A$ and $B$ is
1 $\frac{18}{5} \Omega$
2 $\frac{16}{3} \Omega$
3 $3 \Omega$
4 $5 \Omega$
Explanation:
A Given circuit, Circuit is balanced by wheat stone bridge so no current flow in centre. $\mathrm{R}_{\mathrm{AB}}=\frac{6 \times 9}{9+6}=\frac{54}{15}=\frac{18}{5} \Omega$
