A 9 ohm resistor are connected in series to a power supply. The voltage drop acr

Question

A 9 ohm resistor are connected in series to a power supply. The voltage drop across the 6 ohm resistor is measured with a amphmeter and found to be 12.0 volts. a. draw the circuit diagram including the power properly connected to the circuit to make this measurement. b. what is the cureent through the circuit. c. what is the voltage drop over the 9 ohm resistor? The same two 9 ohm and 6 ohm resistor are now connected in parallel to the power supply, but the voltage setting is damaged on the power supply. the current through the 9 ohm resistor is measured with an ammeter and found to be 0.25 amp. a. draw the circuit for this arrangement, including the power supply and the ammeter (measures current) correctly connected to make this measurement. b. what is the voltage setting of the power supply? c. what is the current through the 6 ohm resistor with this new arrangement?

Explanation / Answer

V=IR 12=Ix6 Current In the circuit =2A Voltage drop across 9ohm =9x2=18 V 2nd Case Voltage of power supply =0.25x9 =2.25V Current through 6ohm =2.25/6 =0.375A

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