A 500 m long wire has a resistance of 20 Ω. How much must be cut off to reduce its resistance to 12 Ω?
Given quantities:
l=500m R=20Ω R1=12Ω Δl−?l = 500m \space\space\space R = 20\varOmega \space\space\space R_1 = 12\varOmega \space\space\space\space\space\space \Delta l -?l=500m R=20Ω R1=12Ω Δl−?
R=ρlS R1=ρl1SR = \rho\frac{l}{S} \space\space\space\space\space\space R_1 = \rho\frac{l_1}{S}R=ρSl R1=ρSl1
RR1=ll1→l1=R1Rl=1220∗500=\large\frac{R}{R_1} = \frac{l}{l_1} \to l_1 = \frac{R_1}{R}l=\frac{12}{20}*500 =R1R=l1l→l1=RR1l=2012∗500= 300m300m300m
Δ=l−l!=200m\Delta = l - l_! = 200mΔ=l−l!=200m
The 200m wire is cut to reduce the resistance to 12Ω\varOmegaΩ
Answer: 200m
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