CATAGORY: Electricity & Magnetism
QUESTION: A battery of four 1.5-V dry cells in a series is connected in
series with a 5.0-ohms resistor and a pair of 10.0-ohms resistors in
parallel. (a)Draw the circuit diagram. (b)If the internal resistance of
the battery is negligible, what is the current through one of the
10.0-ohms resistors?

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      __________
1.5V |          |
  ___|___      <
    ---          > 5.0ohm
     |         <
     |      ____|____
     |     |         |
     |    <         <
     |     >10.0ohm   > 10.0ohm
     |    <         <
     |     |_________|
     |          |
     |__________| 
     
     Adding the resistors in parallel gives
     
     1       1            1         2         1
(1)  - =  -------  +   -------  = ------- =  ------
     R    10.0ohm      10.0ohm    10.0ohm    5.0ohm
     
     so the equivelant R is
     
     R = 5.0 ohm
     
     placing this resistor in series with the 5.0 ohm resistor
     gives an equivelant resistance of the whole circuit as
     
(2)  R(equ) = R + 5.0 ohm = 5.0 ohm + 5.0 ohm = 10.0 ohm

the current in the circuit is then

(3) I = V/r(equ) = 1.5V/10.0 ohm = .15 A

the voltage drop across the 5.0 ohm resistor is then

(4) V(5.0 ohm) = I 5.0 ohm = .15 A x 5.0 ohm = .75 V

the voltage drop across each parallel 10.0 ohm component is then

(5) V(10 ohm) = 1.5 V - .75V = .75 V

The current through a 10.0 ohm resistor is then

(6)  I(10.0 ohm) = V(10 ohm)/10.0ohm = .75V /10 =.075A

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*                      *
* I(10.0 ohm) = .075A  *
*                      *
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