# How is the voltage in this parallel circuit different across each component?

I'm studying up on circuit calculations and came across this circuit: I was taught that voltage in a parallel circuit is the same across all components, which would be the voltage supplied by the cells. So why is the reading across the resistor 4V? Shouldn't it be 9V, or is it treated differently because the components within the branch are in series with each other? So could I say the voltage is the same across each branch instead, with different components having different voltages across them, due to their differing resistances?

If so, am I right in saying V1= 9V and V2= 5V?

• That resistor is in series with something else, right? Do you understand why elements actually in parallel have the same voltage drop? Nov 16, 2019 at 19:41
• Is "A1" an ideal ammeter? Nov 16, 2019 at 21:11
• It is not in parallel connection!! And Yes! V1 = 9V and V2= 5V. 5V because it is in series with 4V and the total voltage across them is 9V so what you got left is 5V.
– user316791
Jan 24, 2022 at 19:02

In parallel circuit voltage is the same among branches but current is different which you can calculate using Ohms law: $$U=I\cdot R$$

In series current stays the same but voltage drops proportionally per resistance. The voltage drop can also be calculated using Ohms law.

In this example, yes - V2 will show 5V and V1 will show 9V.

Using Ohms law we can calculate that in the lower part of the circuit the total resistance is 4,5 Ohms as 2 Amperes are flowing through it.

In series total resistance is calculated like this: $$R=R_1+R_2$$

Thus we can further calculate that the resistor has a resistance of 2 Ohms and the lower light bulbs resistance is 2,5 Ohms.

In parallel circuit the total resistance is calculated like this: $$\frac1R=\frac1{R_1}+\frac1{R_2}$$

Then using Ohms law again we can calculate the total current flowing throw this circuit which would be shown in A1. The total resistance of circuit is 1,8 Ohms so total current is 5 Amperes.