1. A 9.0-Ω resistor and a 6.0-Ω resistor are connected in series with a power supply. (a) The voltage drop across the 6.0-Ω resistor is measured to be 12 V. Find the voltage output of the power supply. (b) The two resistors are connected in parallel across a power supply, and the current through the 9.0-Ω resistor is found to be 0.25 A. Find the voltage setting of the power supply.
Answers
Part (a)
Given :
Two resistors of 6 ohm and 9 ohm .
Voltage across 6 ohm resistor = 12V
To find :
The voltage output of the power supply .
Solution :
Current in 6 ohm resistor = 12 / 6 = 2 A
As the resistors are in series , the current will be the same in the whole circuit .
Now ,
Net voltage = current * equivalent resistance
=> Net voltage = 2 * ( 6 + 9 )
=> Net voltage = 2 * 15 = 30 V
The voltage output of the power supply is 30 V .
Part (b)
Given :
Two resistors of 6 ohm and 9 ohm .
Current across 9 ohm resistor = 0.25 A
To find :
The voltage output of the power supply .
Solution :
Voltage in 9 ohm resistor = 0.25 * 9 = 2.25 V
As the resistors are in parallel , the voltage will be the same in all the branches equal to the voltage of the power supply .
Now ,
Net voltage = Voltage in 9 ohm resistor = 2.25 V
The voltage output of the power supply is 2.25 V .