A) No charge
B) Brilliance of \[{{L}_{1}}\]decreases and that of \[{{L}_{2}}\]increases
C) Brilliance of both \[{{L}_{1}}\]and \[{{L}_{2}}\]increases
D) Brilliance of both \[{{L}_{1}}\]and \[{{L}_{2}}\]decreases
Correct Answer: B
Solution :
Let R be the resistance of each lamp. If E be the applied e. m. f. then the current in the circuit I, is given by \[{{I}_{1}}=\frac{E}{R+(R/2)}=\frac{2E}{3R}\] Current flowing through \[{{L}_{2}}\] or \[{{L}_{2}}\] When\[{{L}_{3}}\]is fused, the whole current flows through\[{{L}_{1}}\] and \[{{L}_{1}}.\]Thus \[{{I}_{2}}=\frac{E}{2R}\] So, current through \[{{L}_{1}}\]decreases and through \[{{L}_{2}}\]increases. Hence, the correction option is [b].You need to login to perform this action.
You will be redirected in
3 sec