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an 11 ohm resistor and a 6 ohm resistor are connected in series with a battery. the potential difference across the 6 ohm resistor is 9 volts. find the potential difference across the battery

Sagot :

AL2006

Two resistors in series are often called a 'voltage divider', because the
total voltage divides in proportion to the resistances.

The total resistance in the string across the battery is (11 + 6) = 17 ohms.

-- The full battery voltage appears across 17 ohms.
-- The voltage across the 11-ohms is (11/17) of the battery, and
-- the voltage across the 6-ohms is (6/17) of the battery. 

                                  (6/17) x (B) = 9 volts

Multiply each side by (17/6) :      B = (9 volts) x (17/6)  =  25.5 volts .

By the way, in case you care or are asked . . .

-- The current in the whole series loop is  B/R = 25.5 / 17 = 1.5 Amperes
-- The power drawn from the battery is   B²/R = (25.5)²/17 =  38.25 watts
-- The power dissipated by the 6-ohm resistor is  V²/R = 9²/6 = 13.5 watts
-- The power dissipated by the 11-ohm resistor is I²R = (1.5)² (11) = 24.75W
-- (Check:  13.5W + 24.75W = 38.25W     yay! )
-- If they're just composition units hanging out in the air, then both resistors
are getting quite warm.


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