MHT CET 2025 19th April Evening Shift | Current Electricity Question 9 | Physics

Two cells $E_1$ and $E_2$ having equal e.m.f ' $E$ ' and internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively are connected in series. This combination is connected to an external resistance ' R '. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

$\mathrm{r}_1-\mathrm{r}_2$

$r_1+r_2$

$\frac{r_1-r_2}{2}$

$\frac{r_1+r_2}{2}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

If only $5 \%$ of the total current is to be passed through galvanometer of resistance G , then the resistance of the shunt will be

$\frac{\mathrm{G}}{15}$

$\frac{\mathrm{G}}{17}$

$\frac{\mathrm{G}}{19}$

$\frac{\mathrm{G}}{21}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of 16 mA and maximum voltage is 80 mV . Now the range is changed so that galvanometer reads 160 V . The required resistance to be connected is

$9995 \Omega$ in series.

$4995 \Omega$ in series.

$9.5 \times 10^{-3} \Omega$ in parallel.

$4.95 \times 10^{-3} \Omega$ in parallel.

MHT CET 2024 16th May Evening Shift

MCQ (Single Correct Answer)

-0

Two identical galvanometers are converted into an ammeter and into milliammeter. For the same current, the value of shunt of the ammeter as compared to that of milliammeter is

less

equal

zero

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