Question

#52 A simple ohmmeter is made by connecting a 1.50V flash- light battery in series with a resistance R and an ammeter that

please complete each section of the problem:a,b,c,d

Answer 1

When the clips are shorted together

$V=IR$

$\Rightarrow 1.5=1\times 10^{-3}\times R$

$\Rightarrow R=\frac{1.5}{1\times 10^{-3}}=1500\ \Omega$

So, the value of R is 1500 Ohm.

PART A: When Current is 10% of its maximum value of 1.5 mA.

$V=I(R+r)$

$1.5=\frac{1\times 10^{-3}\times 10}{100}(1500+r)$

$\Rightarrow 1500+r=15000$

$\Rightarrow r=15000-1500=13500\ \Omega$

So, for 10% of max current, the resistance should be 13500 Ohm.

PART B: When Current is 50% of its maximum value of 1.5 mA.

$V=I(R+r)$

$1.5=\frac{1\times 10^{-3}\times 50}{100}(1500+r)$

$\Rightarrow 1500+r=3000$

$\Rightarrow r=3000-1500=1500\ \Omega$

So, for 50% of max current, the resistance should be 1500 Ohm.

PART C: When Current is 90% of its maximum value of 1.5 mA.

$V=I(R+r)$

$1.5=\frac{1\times 10^{-3}\times 90}{100}(1500+r)$

$\Rightarrow 1500+r=1667$

$\Rightarrow r=1667-1500=167\ \Omega$

So, for 90% of max current, the resistance should be 167 Ohm.

PART D:

If the ammeter has an internal resistance of 20 ohms. Then

$V=I(R+r)$

$\Rightarrow 1.5=1\times 10^{-3}\times (R+20)$

$\Rightarrow R=1500-20=1480\ \Omega$

So, if the ammeter has an internal resistance of 20 Ohms, the value of R would be 1480 ohms.