There is only one possible answer.
Voltage = current * resistance
also, Power = Voltage * Voltage / resistance
If you follow the algebra, you should get a resulting resistance of the lightbulb to be 548.6 ohms.
The net resistance of two resistors connected in series is the sum of the two resistances. RSERIES = Summation1toN RN
If the resistors are in series, then the total resistance is simply the sum of the resistances of each resistor.
If the resistors are connected in series, the total resistance will be the sum of the resistances of each resistor, and the current flow will be the same thru all of them. if the resistors are connected in parallel, then the current thru each resistor would depend on the resistance of that resistor, the total resistance would be the inverse of the sum of the inverses of the resistance of each resistor. Total current would depend on the voltage and the total resistance
Basically, your series and shunt ohmmeters differ in circuit configuration. Your series ohmmeter is configured in a way that your "meter" (which has internal resistance) is connected in series to your "measured resistor" and we all know that those TWO resistances will ADD up causing some sort of inaccuracy. While on the other hand, your shunt ohmmeter is configured in a way wherein your "meter" is connected in PARALLEL to your "measured resistor" that will ease-up the inaccuracy but will only measure resistances ranging from 200 Ohms to 400 Ohms (typically and depending on your configuration).
A voltmeter can be connected in parallel with a resistor to show the voltage across the resistor.
The voltage drop across each resistor is determined by the amounts of resistance in the 3 resistors and all the rest of the resistances in the electrical circuit.
forcing a constant current and measuring the voltage across the unknown resistor.
Divide the voltage
Resistances in series act just as if they were one single resistor. The value of the single resistor is the sum of the individual resistors connected in series ... Ra + Rb + Rc etc. When several resistors are in series, the effective total is greater than the biggest one. Resistance in parallel act just as if they were one single resistor. The reciprocal of the value of the single resistor is the sum of the reciprocals of the individual resistors connected in parallel ... Total effective resistance = 1 divided by (1/Ra + 1/Rb + 1/Rc + etc.) When several resistors are in parallel, the effective total is less than the smallest one. Once you figure out the effective value of the series- or parallel-combination of many resistors, you handle them as if they were one single resistor, and you can work with the voltage and current: Current through any resistance = (Voltage across it) divided by (its resistance).
Even though it is connected to a 9 volt source, it is still a resistor.
That completely depends on the voltage of the battery.The energy delivered by the battery and dissipated by the resistor in one minute will be[ 60 x (Voltage of the battery)2 / 21 ] joules
I'd have to see a diagram, as your description in words is unclear.