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P-N Junction

Two ideal bat...

Question

Two ideal batteries of emf $V_{1}$ and $V_{2}$ and resistances $R_{1}$ , $R_{2}$ and $R_{3}$ are connected as shown in the figure. The current in resistance $R_{2}$ would be zero if

A

V_{1} = V_{2} and R_{1} = R_{2} = R_{3}

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B

V_{1} = V_{2} and R_{1} = 2 R_{2} = R_{3}

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C

V_{1} = 2 V_{2} and 2 R_{1} = 2 R_{2} = R_{3}

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D

2 V_{1} = V_{2} and 2 R_{1} = R_{2} = R_{3}

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Solution

The correct options are
A $V_{1} = V_{2} and R_{1} = R_{2} = R_{3}$
B $V_{1} = V_{2} and R_{1} = 2 R_{2} = R_{3}$

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Similar questions

Q. The emf (V) and internal resistance (

r

) of a single battery which is equivalent to a parallel combination of two batteries of emf's

V_{1}

V_{2}

and internal resistance

r_{1}

r_{2}

, respectively, with polarities as shown in the figure are given by

\frac{V_{1} r_{2} - V_{2} r_{1}}{r_{1} + r_{2}}, \frac{r_{1} r_{2}}{r_{1} + r_{2}}

.

Q. Find the current flowing through the resistance

R_{1}

of the circuit shown in figure, if the resistances are equal to

R_{1} = 10 Ω

R_{2} = 10 Ω

R_{3} = 10 Ω

, and the potential of points

1, 2

3

V_{1} = 10 V

V_{2} = 6 V

V_{3} = 5 V

r_{1} < r_{2} < r_{3}

are the ex-radii of a right angled triangle and

r_{1} = 1, r_{2} = 2

r_{3} = . . . .

Q. Assertion :In a

△ A B C

a < b < c

r

is inradius and

r_{1}, r_{2}, r_{3}

are the axradii opposite to angle

A, B, C

respectively, then

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}

is equal to
(where

r_{1}, r_{2}

r_{3}

are the exradii and

2 s

is the perimeter.)

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