Electricity doubt
I tried using KCL and Ohm's Law but still couldn't solve this. Please give me a detailed solution for this question.
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@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.total resistance = 10 Ω
total V = 10 V
current = V/R = 1 A
since V=iR
so about the 4 Ω resistor the voltage drop would be 1×4 = 4 V
so the voltmeter reading would be 4 V
How can it be 10 ohm? We have to consider resistance of V too right? If not, then why?
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Requested by imthefool15 | User ID: 1294332914850336878
Yo please don't mind that it was accidental
V tends to infinity so we don’t consider it that is why volt meters are connected parallel
1/4 + 1/∞ =1/R_net
See how it will just come as 1/4
In some cases they may mention the resistance of volt meter then just use the parallel addition formula
Req = 6+4 = 10 ohms
V=10
I =1A by ohms law
potential difference across 4 ohm resistor = current flowing through circuit *4 ohm
V' = 1×4=4V
ur in fiitjee right? as i have the same modules
awesome explanation.
Thank you everyone
Yeah
+solved @727 @waste of space @iq21dumbaahh @iTeachChem
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