Question

a new flashlight cell of emf 1.5 volts gives a current of 15amps, when connected directly to an ammemeter of resistance 0.04. the internal resistance of cell is​

Answer 1

• I=E/(R+r)

I=CURRENT

E=EMF OF CELL

R=RESISTANCE

r=INTERNAL RESISTANCE

• 15=1.5/(0.04+r)

NOW MULTIPLY BOTH SIDES BY (0.04+r)

• 15×(0.04+r)=1.5
• 60/100 + 15r=1.5
• 3/5 + 15r=1.5
• (3 + 75r)/5=1.5
• 3+75r=7.5
• 75r=7.5-3
• 75r=4.5
• r=4.5/75
• r=0.06

THEREFORE INTERNAL RESISTANCE OF THE CELL IS 0.06