username5288572
Badges: 10
Rep:
?
#1
Report Thread starter 2 weeks ago
#1
Could someone explain how to do 1bName:  IMG_20210421_013828.jpg
Views: 12
Size:  126.2 KB
Last edited by username5288572; 2 weeks ago
0
reply
lordaxil
Badges: 9
Rep:
?
#2
Report 2 weeks ago
#2
To work it out, remember that current through 4.7 kΩ resistor (I_1) and analogue voltmeter (I_2) add to give the total current in circuit (I) which you worked out already in part (a). Then, work out I_1 from the voltage (3.2 V) given across 4.7 kΩ resistor, allowing you to work out I_2. Finally, you need the voltage drop across analogue voltmeter, which must be the original voltage drop in part (a) minus the voltage drop across 4.7 kΩ resistor when the analogue voltmeter is present.

(if the part sentence is not obvious, think about the analogue voltmeter as "leaking" some current around the 4.7 kΩ resistor, so the voltages add in series around the short circuit)
Last edited by lordaxil; 2 weeks ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you haven't confirmed your firm and insurance choices yet, why is that?

I don't want to decide until I've received all my offers (125)
35.41%
I am waiting until the deadline in case anything in my life changes (47)
13.31%
I am waiting until the deadline in case something in the world changes (ie. pandemic-related) (22)
6.23%
I am waiting until I can see the unis in person (35)
9.92%
I still have more questions before I made my decision (46)
13.03%
No reason, just haven't entered it yet (41)
11.61%
Something else (let us know in the thread!) (37)
10.48%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise