What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm
1 answer:
<h3><u>Given</u> :</h3>
Current flow light bulb = 2.5
Resistance of light bulb = 3.6Ω
<h3><u>To Find </u>:</h3>We have to find voltage of battery
➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.
➝ V ∝ I
➝ <u>V = I × R</u>
Where, R is the resistance of conductor.
⇒ V = I × R
⇒ V = 2.5 × 3.6
⇒ <u>V = 9 volt</u>
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Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =
v₂ = 19*
v₂ = 25.8999
v₂ ≈ 26 L Option b.
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =
v₂ = 19*
v₂ = 25.8999
v₂ ≈ 26 L Option b.