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IP Three resistors, R1 = 15 Ω , R2 = 62 Ω , and R3=R, are...

IP Three resistors, R1 = 15 Ω , R2 = 62 Ω , and R3=R, are connected in series with a 24.0 V battery. The total current flowing through the battery is 0.15 A . Find the value of resistance R. Find the potential difference across each resistor.
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Answer #1

Solution- From the question we have
    R1 = 15 Ω :R2 = 62 Ω :R3   =R
    voltage   V = 24 V
     current   I =0.15A
       The three capacitors are connected in series combination  
I = 24/15+62+R

=>0.15 = 24/15+62+R

=>R = 24- 0.15(15+62)/(0.15)
=83 Ω

b) V(15)= 0.15*15= 2.25 V

V(62)= 0.15*62=9.3 V

V(83)= 0.15*83= 12.45 V


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