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A galvanometer of resistance `70 Omega`, is converted to an ammeter by a shunt resistance `r_(3) = 0.03 Omega`. The value of its resistance will becomeA. `0.025 Omega`B. `0.022Omega`C. `0.035Omega`D. `0.030Omega` |
Answer» Correct Answer - D `R=(R_(G)r_(s))/(R_(G) + r_(s))` Here, `R_(G) = 70 Omega, r_(s) = 0.03 Omega` `therefore R=(70 xx 0.03)/(70 +0.03) = 0.02998 = 0.03 Omega` |
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