MCQ
The current from the battery in circuit diagram shown is .............. $A$


- ✓$1$
- B$2$
- C$1.5$
- D$3$

On further solving equivalent resistance $R = 15\,\Omega $
Hence current from the battery $i = \frac{{15}}{{15}} = 1\,A$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $I$ | List $II$ |
| $P.$ $\quad$ Alpha decay | $1.$ $\quad{ }_8^{15} 0 \rightarrow{ }_7^{15} N +\ldots \ldots$. |
| $Q.$ $\quad$ $\beta^{+}$decay | $2.$ $\quad{ }_{92}^{238} U \rightarrow{ }_{90}^{234} Th +\ldots \ldots$. |
| $R.$ $\quad$Fission | $3.$ $\quad{ }_{83}^{185} Bi \rightarrow{ }_{82}^{184} Pb +\ldots \ldots$. |
| $S.$ $\quad$Proton emission | $4.$ $\quad{ }_{94}^{239} Pu \rightarrow{ }_{57}^{140} La +\ldots \ldots$. |
Codes: $ \quad \quad P \quad Q \quad R \quad S $


$\oint {\vec B} .d\vec A\, = \,0$