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 3.27 V. When the switch is closed, the voltmeter reading drops to 3.09 V and the ammeter (A) reads 0.154 A.

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

How much power is being delivered to the load?

How much power is being lost to the internal resistance?

Explanation / Answer

writing the equations,

E=3.33V

again E=i(r+R)

or and E-ir=3.09

or 3.27=0.154*(r+R)

and 3.09=3.27-0.154*r

solving we get r=1.17 ohm

and R=20.06 ohm

A)EMF=3.27V

B)PD across R=0.154*20.06

=3.09 V

so %=3.09/3.27 *100

=94.49%

C)internal resistance r=1.17 ohm

D)R=20.06 ohm

E)Power=I^2R

=0.475W

F)P=i^2r

=0.0277W

G)b) The amount of voltage lost goes down.

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.