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total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.
Current, by Ohm's Law, is voltage divided by resistance, so, 18 volts divided by 2 ohms is 9 amperes.
5 ohms in parallel with 20 ohms is 4 ohms. 4 ohms across 200 volts is 50 amperes. However, resistance is a function of temperature, so the 4 ohms will probably be higher, reducing the current. How much depends on the temperature coefficient of the loads.
Taking the question at face value, the internal resistances will be treated like "real" resistors in the circuit. That means we have 3 batteries of 1.5 volts each connected in series with their 2 ohms + 2 ohms + 2 ohms of internal risistance, or 6 ohms of internal resistance. The 6 ohms of internal resistance acts in series with the 44 ohms of resistance stated as the value of the resistor. The total resistance is simply the sum of the two, or 6 ohms + 44 ohms or 50 ohms of total resistance. The batteries are connected in series, and their individual voltages are added to find total applied voltage. That means 1.5 volts + 1.5 volts + 1.5 volts or 4.5 volts will be the total applied voltage. Total current in the circuit (and through our 44 ohm resistor) will be the voltage applied divided by the resistance ( I = E / R), which, in this case, is 4.5 volts / 50 ohms which equals 0.07 amps. That's 7/100ths of an amp, or, in electronics speak, 70/1000ths of an amp, or 70 milliamps, or 70mA of current. The circuit is a series circuit, and that current, the total circuit current, will be flowing through each and every component of the circuit. That's what a series circuit means.
Assuming you are talking about an AC circuit, then the total opposition to the flow of current in an R-C circuit is called its impedance (symbol: Z), measured in ohms. This is the vector sum of the circuit's resistance (R) and its capacitive reactance (XC) -each also measured in ohms.
total voltage = 4.5V, total resistance = 3.5 ohms, loop current = 4.5V / 3.5 ohms = 1.286Atotal voltage = 9V, total resistance = 4 ohms, loop current = 9V / 4 ohms = 2.25Atotal voltage = 13.5V, total resistance = 4.5 ohms, loop current = 13.5V / 4.5 ohms = 3Aetc.There is no solution to your problem conditions.
Current, by Ohm's Law, is voltage divided by resistance, so, 18 volts divided by 2 ohms is 9 amperes.
Impedance is measured in Ohms defined as total resistance to current (Amps) flow
5 ohms in parallel with 20 ohms is 4 ohms. 4 ohms across 200 volts is 50 amperes. However, resistance is a function of temperature, so the 4 ohms will probably be higher, reducing the current. How much depends on the temperature coefficient of the loads.
ohms law calculation for a series circuit - Total Resistance = Total Voltage divided by Total Current
Taking the question at face value, the internal resistances will be treated like "real" resistors in the circuit. That means we have 3 batteries of 1.5 volts each connected in series with their 2 ohms + 2 ohms + 2 ohms of internal risistance, or 6 ohms of internal resistance. The 6 ohms of internal resistance acts in series with the 44 ohms of resistance stated as the value of the resistor. The total resistance is simply the sum of the two, or 6 ohms + 44 ohms or 50 ohms of total resistance. The batteries are connected in series, and their individual voltages are added to find total applied voltage. That means 1.5 volts + 1.5 volts + 1.5 volts or 4.5 volts will be the total applied voltage. Total current in the circuit (and through our 44 ohm resistor) will be the voltage applied divided by the resistance ( I = E / R), which, in this case, is 4.5 volts / 50 ohms which equals 0.07 amps. That's 7/100ths of an amp, or, in electronics speak, 70/1000ths of an amp, or 70 milliamps, or 70mA of current. The circuit is a series circuit, and that current, the total circuit current, will be flowing through each and every component of the circuit. That's what a series circuit means.
1.5 volts
35 ohms
Assuming you are talking about an AC circuit, then the total opposition to the flow of current in an R-C circuit is called its impedance (symbol: Z), measured in ohms. This is the vector sum of the circuit's resistance (R) and its capacitive reactance (XC) -each also measured in ohms.
Current
Could it be.... 20 ohms(?)
The current in the circuit will depend on how the three resistors are wired. Series? Parallel? Series parallel? With the resistors in series, 3, 2 and 4 ohms will add to 9 ohms. As I = E/R, I = 9 V / 9 ohms = 1 A. With the resistors in parallel, the 3, 2 and 4 ohm resistors will draw 3 A, 4.5 A and 2.25 A respectively, and the total current will be the sum of the branch currents, or 3 A + 4.5 A + 2.25 A = 9.75 A. There are 3 different series parallel circuits possible, and more investigation will be necessary to solve for them.