151801
If an electron revolves in the path of a circle of radius of $0.5 \times 10^{-10} \mathrm{~m}$ at a frequency of $5 \times 10^{15}$ cycle/sec. The electric current in the circle is: (Charge of an electron is $1.6 \times 10^{-19} \mathrm{C}$ )
151802
A $6 \mathrm{~V}$ cell with $0.5 \Omega$ internal resistance, a $10 \mathrm{~V}$ cell with $1 \Omega$ internal resistance and a $12 \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \mathrm{~V}$ cell is
1 0.60
2 2.27
3 2.87
4 5.14
Explanation:
C Applying Kirchhoff's voltage law, $6-0.5 \mathrm{i}_{1}=10-\mathrm{i}_{2}=12\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)$ $6-0.5 \mathrm{i}_{1}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $6=12.5 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $\mathrm{i}_{1}=\frac{6-12 \mathrm{i}_{2}}{12.5}$ And $\quad 10-\mathrm{i}_{2}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $10=12 \mathrm{i}_{1}+13 \mathrm{i}_{2}$ Putting the value of equation (i) in equation (ii), we get $10=12\left(\frac{6-12 \mathrm{i}_{2}}{12.5}\right)+13 \mathrm{i}_{2}$ $10=\frac{72-144 \mathrm{i}_{2}+162.5 \mathrm{i}_{2}}{12.5}$ $18.5 \mathrm{i}_{2}+72=125$ $\mathrm{i}_{2}=\frac{53}{18.5}$ $\mathrm{i}_{2}=2.87 \mathrm{~A}$
EAMCET-2005
Current Electricity
151714
The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to
151715
A current of $0.6 \mathrm{~A}$ is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151801
If an electron revolves in the path of a circle of radius of $0.5 \times 10^{-10} \mathrm{~m}$ at a frequency of $5 \times 10^{15}$ cycle/sec. The electric current in the circle is: (Charge of an electron is $1.6 \times 10^{-19} \mathrm{C}$ )
151802
A $6 \mathrm{~V}$ cell with $0.5 \Omega$ internal resistance, a $10 \mathrm{~V}$ cell with $1 \Omega$ internal resistance and a $12 \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \mathrm{~V}$ cell is
1 0.60
2 2.27
3 2.87
4 5.14
Explanation:
C Applying Kirchhoff's voltage law, $6-0.5 \mathrm{i}_{1}=10-\mathrm{i}_{2}=12\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)$ $6-0.5 \mathrm{i}_{1}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $6=12.5 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $\mathrm{i}_{1}=\frac{6-12 \mathrm{i}_{2}}{12.5}$ And $\quad 10-\mathrm{i}_{2}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $10=12 \mathrm{i}_{1}+13 \mathrm{i}_{2}$ Putting the value of equation (i) in equation (ii), we get $10=12\left(\frac{6-12 \mathrm{i}_{2}}{12.5}\right)+13 \mathrm{i}_{2}$ $10=\frac{72-144 \mathrm{i}_{2}+162.5 \mathrm{i}_{2}}{12.5}$ $18.5 \mathrm{i}_{2}+72=125$ $\mathrm{i}_{2}=\frac{53}{18.5}$ $\mathrm{i}_{2}=2.87 \mathrm{~A}$
EAMCET-2005
Current Electricity
151714
The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to
151715
A current of $0.6 \mathrm{~A}$ is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?
151801
If an electron revolves in the path of a circle of radius of $0.5 \times 10^{-10} \mathrm{~m}$ at a frequency of $5 \times 10^{15}$ cycle/sec. The electric current in the circle is: (Charge of an electron is $1.6 \times 10^{-19} \mathrm{C}$ )
151802
A $6 \mathrm{~V}$ cell with $0.5 \Omega$ internal resistance, a $10 \mathrm{~V}$ cell with $1 \Omega$ internal resistance and a $12 \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \mathrm{~V}$ cell is
1 0.60
2 2.27
3 2.87
4 5.14
Explanation:
C Applying Kirchhoff's voltage law, $6-0.5 \mathrm{i}_{1}=10-\mathrm{i}_{2}=12\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)$ $6-0.5 \mathrm{i}_{1}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $6=12.5 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $\mathrm{i}_{1}=\frac{6-12 \mathrm{i}_{2}}{12.5}$ And $\quad 10-\mathrm{i}_{2}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $10=12 \mathrm{i}_{1}+13 \mathrm{i}_{2}$ Putting the value of equation (i) in equation (ii), we get $10=12\left(\frac{6-12 \mathrm{i}_{2}}{12.5}\right)+13 \mathrm{i}_{2}$ $10=\frac{72-144 \mathrm{i}_{2}+162.5 \mathrm{i}_{2}}{12.5}$ $18.5 \mathrm{i}_{2}+72=125$ $\mathrm{i}_{2}=\frac{53}{18.5}$ $\mathrm{i}_{2}=2.87 \mathrm{~A}$
EAMCET-2005
Current Electricity
151714
The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to
151715
A current of $0.6 \mathrm{~A}$ is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?
151801
If an electron revolves in the path of a circle of radius of $0.5 \times 10^{-10} \mathrm{~m}$ at a frequency of $5 \times 10^{15}$ cycle/sec. The electric current in the circle is: (Charge of an electron is $1.6 \times 10^{-19} \mathrm{C}$ )
151802
A $6 \mathrm{~V}$ cell with $0.5 \Omega$ internal resistance, a $10 \mathrm{~V}$ cell with $1 \Omega$ internal resistance and a $12 \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \mathrm{~V}$ cell is
1 0.60
2 2.27
3 2.87
4 5.14
Explanation:
C Applying Kirchhoff's voltage law, $6-0.5 \mathrm{i}_{1}=10-\mathrm{i}_{2}=12\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)$ $6-0.5 \mathrm{i}_{1}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $6=12.5 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $\mathrm{i}_{1}=\frac{6-12 \mathrm{i}_{2}}{12.5}$ And $\quad 10-\mathrm{i}_{2}=12 \mathrm{i}_{1}+12 \mathrm{i}_{2}$ $10=12 \mathrm{i}_{1}+13 \mathrm{i}_{2}$ Putting the value of equation (i) in equation (ii), we get $10=12\left(\frac{6-12 \mathrm{i}_{2}}{12.5}\right)+13 \mathrm{i}_{2}$ $10=\frac{72-144 \mathrm{i}_{2}+162.5 \mathrm{i}_{2}}{12.5}$ $18.5 \mathrm{i}_{2}+72=125$ $\mathrm{i}_{2}=\frac{53}{18.5}$ $\mathrm{i}_{2}=2.87 \mathrm{~A}$
EAMCET-2005
Current Electricity
151714
The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to
151715
A current of $0.6 \mathrm{~A}$ is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?
