Question

0.125 C of charge flow out of a

5.00 V battery in 6.30 s. How
much resistance is attached to the
battery?

Answer 1

`\huge \mathrm{\longmapsto \: Answer}`

• Charge (q) = 0.125 C

• Time (t) = 6.30 sec

• Potential difference (v) = 5.00 V

• electric current (i) = ?

• resistance (r) = ?

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We know,

`\boxed{\mathrm{current = (charge)/(time)} }`

• 
  `i = (0.125)/(6.30)`

• 
  `i = (125)/(6300)`

Now,

`\boxed{ \mathrm{potential \: \: difference = i* r}}`

• 
  `5 = (125)/(6300) * r`

• 
  `r = (5 * 6300)/(125)`

• 
  `r = (6300)/(25)`

• 
  `\boxed{ \mathrm {r = 252 \: \: ohms}}`

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`\mathrm{ \#TeeNForeveR}`