# Electricity (probably easy

1. Jan 14, 2008

### dumb__x

[SOLVED] Electricity (probably easy

1. The problem statement, all variables and given/known data

In a circuit shown, the p.d. across the 4Ω resistor is 8 V.
a)What is the current through the 4Ω resistor?
b)What is the current through the 3Ω resistor?
c)What is the p.d. across the 3Ω resistor?
d)What is the p.d. applied by the battery?

2. Relevant equations
Ohms Law V=IR

3. The attempt at a solution

a)2A?

#### Attached Files:

• ###### Electricity.jpg
File size:
8.5 KB
Views:
55
2. Jan 14, 2008

### arunbg

Please show how you arrived at the solution. Since the attachment is pending approval from moderators, I can't see your circuit. A quicker way would be to upload the attachment to an online image hosting site like http://photobucket.com" [Broken] and use the corresponding link in IMG tags while posting. Can you do that?

Last edited by a moderator: May 3, 2017
3. Jan 14, 2008

### dumb__x

Can you see it?
<a href="http://s262.photobucket.com/albums/ii120/reika_xx/?action=view&current=Electricity.jpg" [Broken] target="_blank"><img src="http://i262.photobucket.com/albums/ii120/reika_xx/Electricity.jpg" [Broken] border="0" alt="Electricity Circuit"></a>

a)V=IR
8=I*4
I=2A

Last edited by a moderator: May 3, 2017
4. Jan 14, 2008

### dumb__x

5. Jan 14, 2008

### rohanprabhu

1
i. For the 4$\Omega$ resistor,
$$V = 8~NmC^{-1}$$
$$R = 4~NmC^{-2}s^{-1}$$

Hence,

$$I = \frac{V}{R} = \frac{8}{4} \frac{NmC^{-1}}{NmC^{-2}s^{-1}}$$
$$I = 2~Cs^{-1} = 2~A$$

ii. Current through the 3$\Omega$ resistor is the same as through the 4$\Omega$ resistor as both of them are in series.

iii. For the 3$\Omega$ resistor:

$$I = 2~Cs^{-1}$$
$$R = 3~NmC^{-2}s^{-1}$$

Hence,

$$V~=~IR = 2 \times 3~Cs^{-1}.NmC^{-2}s^{-1}$$
$$V = 6~NmC^{-1} = 6~V$$

iv. Do you mean the potential difference across the terminals of the battery? If that is the case, you can solve it using Kirchhoff's law. Refer to the figure in the attachment.

Hence,

$$\epsilon + 2(4)~NmC^{-1} + 2(3)~NmC^{-1} = 0$$
$$|\epsilon| = 14~NmC^{-1} = 14~V$$

So, the battery provides and e.m.f of 14V. [In case you can't see the attachment, go here: http://image.bayimg.com/iajbbaaba.jpg [Broken]]

#### Attached Files:

• ###### circuit1.jpg
File size:
2 KB
Views:
49
Last edited by a moderator: May 3, 2017
6. Jan 14, 2008

### dumb__x

thank you so much!

7. Jan 14, 2008

### fermio

I=U/R=8V/4om=2A
I=8V/3om=8/3 A

This questions I don't understand
"c)What is the p.d. across the 3Ω resistor?
d)What is the p.d. applied by the battery?"

"sorry
http://s262.photobucket.com/albums/i...lectricity.jpg" [Broken]
I=8V/(4om+3om)=8/7 A.
If electricity is ~, then (8/7)A/2=8/14 A, becouse of diod.

Last edited by a moderator: May 3, 2017
8. Jan 14, 2008

### rohanprabhu

@fermio: I am not able what you are trying to say due to poor formatting. Please repost with proper formatting and punctuation.