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

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 voltsMultiply 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.