Potential drop across 5 - is 10 V- then EMF of the battery isA- 12 VB- 28 VC- 30 VD- 40 V

If voltage drop across 5 Ω is 10 V, then current in this branch, nbsp; i = Δ vR =105 =2 A for 15 Ω, nbsp; Δ v =i R = (2) (15) =30 V total drop =30+10 = 40 V ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Resistors in Series and Parallel

Potential dro...

Question

Potential drop across $5 Ω$ is $10 V$ , then EMF of the battery is

12 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

28 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $40 V$
If voltage drop across $5 Ω$ is 10 V, then
current in this branch,
$i = \frac{Δ v}{R} = \frac{10}{5} = 2 A$
for $15 Ω$ , $Δ v = i R = (2 A) (15) = 30 V$
total drop =30+10 = 40 V
Hence, EMF = 40 V

Suggest Corrections

Similar questions

Q. Find the potential difference across the resistor, when

125 J

of heat is produced each second in a

5 Ω

resistor.

Q. Potential drop across

5 Ω

10 V

, then EMF of the battery is

Q. Three resistances

2 Ω, 4 Ω, 5 Ω

are combines in series and this combination is connected to a battery of

12

V emf and negligible internal resistance. The potential drop across these resistances are.

Q. If this battery is connected to a 15

Ω

resistor R whose emf is 40V and an internal resistance of 5

Ω

. Calculate the voltage drop across R.

Q. In the given circuit if point

C

is connected to the earth and a potential of

+ 80 V

is given to the point

A

, the potential at

B

Join BYJU'S Learning Program

Potential drop across 5 Ω is 10 V, then EMF of the battery is

The correct option is D 40 VIf voltage drop across 5 Ω is 10 V, then current in this branch, i=ΔvR=105=2 A for 15Ω, Δv=iR=(2A)(15)=30 V total drop =30+10 = 40 V Hence, EMF = 40 V

Potential drop across $5 Ω$ is $10 V$ , then EMF of the battery is

The correct option is D $40 V$
If voltage drop across $5 Ω$ is 10 V, then
current in this branch,
$i = \frac{Δ v}{R} = \frac{10}{5} = 2 A$
for $15 Ω$ , $Δ v = i R = (2 A) (15) = 30 V$
total drop =30+10 = 40 V
Hence, EMF = 40 V