357388
Two batteries are joined to a resistance of \(3\Omega \) as shown in figure. Match Column I with Column II. Column I Column II A Potential drop across A P A B Potential drop across B Q B C Power is supplied by battery R 14 \(V\) D Power is consumed by battery S 11 \(V\)
1 A - Q, B - R, C - S, D - P
2 A - Q, B - S, C - R, D - P
3 A - R, B - S, C - P, D - Q
4 A - P, B - Q, C - R, D - S
Explanation:
Current flowing in the circuit As \({\varepsilon _A} > {\varepsilon _B}\) battery A is discharging and battery B is in charging. From Kirchoff’s law \(15 - 3I - I - 10 - I = 0\) \( \Rightarrow I = \frac{{\left( {15\,V - 10\,V} \right)}}{{\left( {3\Omega + 1\Omega + 1\Omega } \right)}} = 1\;A\) Potential drop across A \(\Delta {V_A} = {\varepsilon _A} - Ir = 15V - \left( {1\Omega \times 1A} \right) = 14\;V\) Potential drop across B \(\Delta {V_B} = {\varepsilon _B} + Ir = 10V + \left( {1\Omega \times 1A} \right) = 11\;V\) Power is supplied by battery A which is consumed by battery B. The correct option is (3).
PHXII03:CURRENT ELECTRICITY
357389
The current \({I_1}\) (in \(A\)) flowing through \(1\Omega \) resistor in the following circuit is -
357388
Two batteries are joined to a resistance of \(3\Omega \) as shown in figure. Match Column I with Column II. Column I Column II A Potential drop across A P A B Potential drop across B Q B C Power is supplied by battery R 14 \(V\) D Power is consumed by battery S 11 \(V\)
1 A - Q, B - R, C - S, D - P
2 A - Q, B - S, C - R, D - P
3 A - R, B - S, C - P, D - Q
4 A - P, B - Q, C - R, D - S
Explanation:
Current flowing in the circuit As \({\varepsilon _A} > {\varepsilon _B}\) battery A is discharging and battery B is in charging. From Kirchoff’s law \(15 - 3I - I - 10 - I = 0\) \( \Rightarrow I = \frac{{\left( {15\,V - 10\,V} \right)}}{{\left( {3\Omega + 1\Omega + 1\Omega } \right)}} = 1\;A\) Potential drop across A \(\Delta {V_A} = {\varepsilon _A} - Ir = 15V - \left( {1\Omega \times 1A} \right) = 14\;V\) Potential drop across B \(\Delta {V_B} = {\varepsilon _B} + Ir = 10V + \left( {1\Omega \times 1A} \right) = 11\;V\) Power is supplied by battery A which is consumed by battery B. The correct option is (3).
PHXII03:CURRENT ELECTRICITY
357389
The current \({I_1}\) (in \(A\)) flowing through \(1\Omega \) resistor in the following circuit is -
357388
Two batteries are joined to a resistance of \(3\Omega \) as shown in figure. Match Column I with Column II. Column I Column II A Potential drop across A P A B Potential drop across B Q B C Power is supplied by battery R 14 \(V\) D Power is consumed by battery S 11 \(V\)
1 A - Q, B - R, C - S, D - P
2 A - Q, B - S, C - R, D - P
3 A - R, B - S, C - P, D - Q
4 A - P, B - Q, C - R, D - S
Explanation:
Current flowing in the circuit As \({\varepsilon _A} > {\varepsilon _B}\) battery A is discharging and battery B is in charging. From Kirchoff’s law \(15 - 3I - I - 10 - I = 0\) \( \Rightarrow I = \frac{{\left( {15\,V - 10\,V} \right)}}{{\left( {3\Omega + 1\Omega + 1\Omega } \right)}} = 1\;A\) Potential drop across A \(\Delta {V_A} = {\varepsilon _A} - Ir = 15V - \left( {1\Omega \times 1A} \right) = 14\;V\) Potential drop across B \(\Delta {V_B} = {\varepsilon _B} + Ir = 10V + \left( {1\Omega \times 1A} \right) = 11\;V\) Power is supplied by battery A which is consumed by battery B. The correct option is (3).
PHXII03:CURRENT ELECTRICITY
357389
The current \({I_1}\) (in \(A\)) flowing through \(1\Omega \) resistor in the following circuit is -
357388
Two batteries are joined to a resistance of \(3\Omega \) as shown in figure. Match Column I with Column II. Column I Column II A Potential drop across A P A B Potential drop across B Q B C Power is supplied by battery R 14 \(V\) D Power is consumed by battery S 11 \(V\)
1 A - Q, B - R, C - S, D - P
2 A - Q, B - S, C - R, D - P
3 A - R, B - S, C - P, D - Q
4 A - P, B - Q, C - R, D - S
Explanation:
Current flowing in the circuit As \({\varepsilon _A} > {\varepsilon _B}\) battery A is discharging and battery B is in charging. From Kirchoff’s law \(15 - 3I - I - 10 - I = 0\) \( \Rightarrow I = \frac{{\left( {15\,V - 10\,V} \right)}}{{\left( {3\Omega + 1\Omega + 1\Omega } \right)}} = 1\;A\) Potential drop across A \(\Delta {V_A} = {\varepsilon _A} - Ir = 15V - \left( {1\Omega \times 1A} \right) = 14\;V\) Potential drop across B \(\Delta {V_B} = {\varepsilon _B} + Ir = 10V + \left( {1\Omega \times 1A} \right) = 11\;V\) Power is supplied by battery A which is consumed by battery B. The correct option is (3).
PHXII03:CURRENT ELECTRICITY
357389
The current \({I_1}\) (in \(A\)) flowing through \(1\Omega \) resistor in the following circuit is -
