For an ideal battery (r = 0 ), closing the switch in (Figure 1) does not affect

Question

For an ideal battery (r = 0 ), closing the switch in (Figure 1) does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 1.50 V battery has an internal resistance r = 0.60 and that the resistance of a glowing bulb is R = 6.00 . 1- What is the current through bulb A when the switch is open? 2-What is the current through bulb A after the switch has closed? 3- By what percent does the current through A change when the switch is closed? CO

Explanation / Answer

a)

when switch is open

Rtotal = total resistance of the circuit = r + R

E = total Voltage

i = current in the circuit = current in the bulb

using ohm's law

E = i (R + r)

i = E/(R + r) = 1.5 /(6 + 0.6) = 0.23 A

2)

when switch is closed

two bulbs are in parallel and their combination is given as

R' = R x R /(R + R) = R/2

Rtotal = total resistance = R/2 + r

i = current coming from the battery

using ohm's law

E = i Rtotal

1.5 = i (6/2 + 0.6)

i = 0.42 A

V = Voltage across the bulb A = i R' = iR/2 = 0.42 x 6/2 = 1.26 volts

ia = current in bulb A = V/R = 1.26/6 = 0.21 A

since the current has dropped , the bulb gets dimmer

3)

%age change = (0.23 - 0.21) x 100 /(0.23) = 8.7%

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