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2 years ago

8

Two identical batteries have the same internal resistance. Battery 1 is connected to a resistor R1 and battery 2 to a resistor R

2 > R1. Compare the terminal voltage of these batteries and the current they provide. A) V1>V2 and I1>I2
B) V1>V2 and I1 C) V1 D) V1I2
E) V1 Please provide explanation

1 answer:

Anna [14]2 years ago

8 0

Answer:

I1 < I2

V1 > V2

Explanation:

Let the EMF of both the batteries is E and the internal resistance of both the cells is r.

The relation between the current and the EMF is given by

$I = \frac{E}{R+r}$

where, R be the resistance connected in the circuit.

the relation between the terminal potential difference and the EMF is given by

V = E - Ir

where, V is the terminal potential difference

If R1 is connected,

The current is given by

$I_{1} = \frac{E}{R_{1}+r}$ .... (1)

The terminal potential difference is given by

$V_{1}=E - I_{1}r$ ..... (2)

If R2 is connected,

The current is given by

$I_{2} = \frac{E}{R_{2}+r}$ .... (3)

The terminal potential difference is given by

$V_{2}=E - I_{2}r$ ..... (4)

As given in the question, R2 > R1

So, by the equation (1) and (3), we get

I1 < I2

and by the equation (2) and (4) ,we get

V1 > V2

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5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

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Answer:

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Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

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in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

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