Question #33648

a dry cell when short-circuited will furnish about 30 amperes for a brief time. If its emf is 1.5 volts, ehat is its internal resistance? an ordinary household electric lamp takes about 1 amp will it be safe to connect it directly to the dry cell? Why?

Expert's answer

I=E/(r+R), where

I current

E EMF, r is internal resistance, R is resistance of whole circuit.

If the circuit is short-circuited then R will be 0 ohm

r=E/i

r=1.5/30=0.05 ohm

P=I*E

P(if R=0)=1.5*30=45 w(on internal resistance)

P(if R not 0, light bulb resistance, for example)=1*1.5=1.5 w - is much less

than 45 w

but P(r)=1*0.05=0.05 w(power on internal resistance, with connected light bulb)

I current

E EMF, r is internal resistance, R is resistance of whole circuit.

If the circuit is short-circuited then R will be 0 ohm

r=E/i

r=1.5/30=0.05 ohm

P=I*E

P(if R=0)=1.5*30=45 w(on internal resistance)

P(if R not 0, light bulb resistance, for example)=1*1.5=1.5 w - is much less

than 45 w

but P(r)=1*0.05=0.05 w(power on internal resistance, with connected light bulb)

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