Batteries are not perfect. They can\'t deliver infinite current. As the current

Question

Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller. We can represent this by treating batteries as if they have some small internal resistance.

The circuit below shows a battery hooked up to a resistor, a voltmeter (for measuring voltages), and an ammeter (for measuring currents).

When you put a measuring device on something, like a circuit, you don't want to change the circuit. So this means that ideal voltmeters have infinite resistance, and ideal ammeters have zero resistance. In reality, neither of these are true, but for this problem, and for all other problems unless you're told otherwise, we'll assume that our ammeters and voltmeters are ideal.

When switch S is open, the voltmeter (V) reads 3.30 V. When the switch is closed, the voltmeter reading drops to 3.12 V and the ammeter (A) reads 0.149 A.

a) Given the available information, what is the emf of the battery?

b) What percent of the battery's emf is actually reaching resistor R, the load resistor? Enter your answer as a number between 0 and 100 with no percent sign.

c)What is the internal resistance 'r' of the battery?

d)What is the resistance R of the load on the circuit? The load resistor represents whatever appliance or device you've hooked up to the circuit.

e)How much power is being delivered to the load?

f)How much power is being lost to the internal resistance?

Explanation / Answer

Given,

E= 3.30 V

again E= i(r+R)

or and E-ir= 3.12

or 3.30 = 0.149 x (r+R)

and 3.12 = 3.30 - 0.149 r

solving we get r= 1.21 ohm

and R= 20.94 ohm

A) EMF= 3.30 V

B) PD across R= 0.149 x 20.94 = 3.12 V

so % = (3.12/3.30 )100 = 94.54 %

C) internal resistance, r= 1.21 ohm

D) R= 20.94 ohm

E) Power = (I^2) R = 0.465 W

F) P= (i^2) r = 0.027 W

Please rate my answer if you find it helpful, good luck...

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