In the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R 1 ?
![part 2 - Question 12 (Choice 1) - CBSE Class 10 Sample Paper for 2022 Boards [Term 2] - Science Solution - Solutions to CBSE Sample Paper - Science Class 10 - Class 10](https://cdn.teachoo.com/47d9c32a-9ee0-4c3f-a507-a08af0b303f6/slide2.jpg)
![part 3 - Question 12 (Choice 1) - CBSE Class 10 Sample Paper for 2022 Boards [Term 2] - Science Solution - Solutions to CBSE Sample Paper - Science Class 10 - Class 10](https://cdn.teachoo.com/bab10d01-0df1-40ab-8208-edb1260ec316/slide3.jpg)
![part 4 - Question 12 (Choice 1) - CBSE Class 10 Sample Paper for 2022 Boards [Term 2] - Science Solution - Solutions to CBSE Sample Paper - Science Class 10 - Class 10](https://cdn.teachoo.com/fce24bf0-951f-498d-8772-f27ddea1245d/slide4.jpg)
CBSE Class 10 Sample Paper for 2022 Boards [Term 2] - Science Solution
CBSE Class 10 Sample Paper for 2022 Boards [Term 2] - Science Solution
Last updated at Dec. 14, 2024 by Teachoo
Transcript
Question 12 (Choice 1) In the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R1? 5 ohm, 10 ohm and R1 are in Parallel 1/𝑅𝑝=1/5+1/10+1/𝑅_1 1/𝑅𝑝=((2+1))/10+1/𝑅_1 =3/10+1/𝑅_1 1/𝑅𝑝=((3 𝑅_1+ 10))/(10 𝑅_1 ) 𝑅𝑝=(10 𝑅_1)/((3 𝑅_1+ 10)) Now, 6 ohm, 6 ohm and Rp are in series Thus, Req = 12 + (10 𝑅_1)/((3 𝑅_1+ 10)) 𝑉=𝐼 Req From the circuit Req = 30/2 = 15 ohms Equation (1) and (2) "12 + " (10 𝑅_1)/((3 𝑅_1+ 10)) "15 + " (10 𝑅_1)/((3 𝑅_1 + 10))=3 10 𝑅_1=(9 𝑅_1+30) Thus, 𝑅_1 = 30 ohm.