MHT CET (PCB) 2024 22th April Evening Shift | Current Electricity Question 1 | Physics

A galvanometer of resistance ' $G$ ' is converted into an ammeter of resistance $\frac{\mathrm{G}}{40}$, by connecting a shunt ' $S$ ' to it. The part of main current passing through the galvanometer is

$25 \%$

$10 \%$

$2 . 5 \%$

$0 . 4 \%$

MHT CET (PCB) 2024 22th April Morning Shift

MCQ (Single Correct Answer)

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Which one of the following is the correct equation for the electric circuit shown in the figure?

MHT CET (PCB) 2024 22th April Morning Shift Physics - Current Electricity Question 3 English

$\quad E_1-\left(i_1+i_2\right) R+i_1 r_1=0$

$\quad E_1-\left(i_1+i_2\right) R-i_1 r_1=0$

$\mathrm{E}_2-\mathrm{i}_2 \mathrm{r}_2-\mathrm{E}_1-\mathrm{i}_1 \mathrm{r}_1=0$

$\quad E_2-\left(i_1+i_2\right) R+i_2 r_2=0$

MHT CET (PCB) 2024 22th April Morning Shift

MCQ (Single Correct Answer)

-0

A galvanometer of resistance ' $G$ ' can be converted into a voltmeter of range ( $0-\mathrm{V}$ ) volt by connecting a resistance ' $R$ ' in series with it. The resistance ' $R$ required to change its range from $\left(0-\frac{\mathrm{V}}{4}\right)$ volt will be

$\frac{R-G}{2}$

$\frac{\mathrm{R}-2 \mathrm{G}}{3}$

$\frac{R-3 G}{4}$

$\frac{4 R-3 G}{5}$

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