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 3.33 V. When the switch is closed, the voltmeter reading drops to 3.20 V and the ammeter (A) reads 0.145 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?

G) As you've seen, not all of a battery's rated voltage makes it to the appliance that the battery is hooked up to. Qualitatively speaking, is this situation made worse, made better, or unaffected if we attach a bigger load resistor? That is, if we make R bigger, does the amount of voltage lost to the internal resistance go up, go down, or stay the same? Note that you only get one try, so look carefully at the equations you've been using to figure out the answer.

Choices for part G)

a) he amount of voltage lost goes up.

b) The amount of voltage lost goes down.
c) The amount of voltage lost stays the same.

Explanation / Answer

writing the equations,

E=3.33V

again E=i(r+R)

or and E-ir=3.2

or 3.33=0.145*(r+R)

and 3.2=3.33-0.145*r

solving we get r=0.8966 ohm

and R=22.069 ohm

A)EMF=3.33 V

B)PD across R=0.145*22.069

=3.2 V

so %=3.2/3.33 *100

=96.096%

C)internal resistance r=0.8966 ohm

D)R=22.069 ohm

E)Power=I^2R

=0.464 W

F)P=i^2r

=0.01885 W

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.