The diagram above shows a meter bridge in which two of thee arms contain resistances R and 2Ω. A balance point is obtained at 60cm from the left end , calculate the value of R

a

1.2Ω

b

1.3Ω

c

3.0Ω

d

6.0Ω

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Richiedebbie
6 years ago

R/2=60/40
cross multiply
R*40=60*2
40R=120
Divide both side by 40
R=3.0

De sage
1 year ago

R1/R2=L1/L2
R1=?
R2=2
L1=60
R2=40

then substitute
R1/2=60/40
R1=60*2/40
R1=120/40
R1=3 ohms

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