If three resistances, 6 ohms, 12 ohms,and 18 ohms, are connected in series across 36 volts,...?

If the resistors are in series, then to find the equivalent resistance you just add up all the resistances. 6 + 12 + 18 = 36 Ohms

To find the current, you take V/R

36 Volts / 36 Ohms = 1 Ampere

Good job on getting the right answer!

Series resistance formula Rt = R1 + R2 + ... + Rn

Total resistance = 6 + 12 + 18 ohms = 36 ohms

Ohm's law: Current = voltage / resistance = 36 / 36 amps

= 1 amp

If the drop in the course of the 8-ohm resistor is 40 8 volts, then with assistance from Ohm's regulation: I = E/R = 40 8/8 = 6 amps by the 8 ohm resistor. === series blend of resistors of 8 and 16 ohms because the resistors are in series, a similar modern-day flows by both one in all them. therefore, on condition that I = E/R and therefore E = I*R, we are able to calculate the voltage in the course of the 16 ohm resistor as E = I*R = 6*16 = ninety six volts. because the 8 ohm resistor is in series the voltages for the time of them upload so the voltage the series blend is E =E(for the time of 16) + E(for the time of 8) = ninety six + 40 8 = one hundred forty four volt. === The 12 ohm resistor is in parallel with the series blend, so the the full voltage for the period of both series resistors is likewise in the course of the 12 ohm resistor. So I(by 12) = E/R = one hundred forty four/12 = 12 amp. === with assistance from Kichoff's modern-day regulation, the full modern-day into the parallel blend is the full modern-day into it truly is branches so I(complete) = I(thrugh12 ohm) + I(by series resistors) = 12 + 6 = 18 amp.

current = voltage /resistance

resistance series =r1 +r2 +r3

current =36/ 12+18 +6= 1 amp

the answer is e) 5,724,094 amp! just kidding, it is b) 1 amp