# In case of circuit breakers, the rate of rise of restriking voltage (RRRV) is expressed in

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In case of circuit breakers, the rate of rise of restriking voltage (RRRV) is expressed in

1. kV/μs
2. μs/​kV
3.  kV
4.  kA/μs

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Correct Answer - Option 1 : kV/μs

The transient voltage appear across the contacts of circuit breaker at current zero period during arcing is known as Restriking voltage.

The expression for the re-striking voltage is ${V_C}\left( t \right) = {V_m}\left( {1 - \cos {ω _n}t} \right)$

The maximum value of the re-striking voltage occurs at  $t = \frac{\pi }{{{ω _n}}} = \pi \sqrt {LC}$

The maximum value of re-striking voltage = 2 Vm = 2 × peak value of the system voltage

The rate of rise of re-striking voltage (RRRV) $= \frac{d}{{dt}}\left[ {{V_m}\left( {1 - \cos {ω _n}t} \right)} \right]$

∴ RRRV = Vm ωn sin ωnt

The rate of rise of re-striking voltage is directly proportional to the natural frequency.

$RRRV \propto {ω _n} \propto \frac{1}{{\sqrt {LC} }}$

Where, L is the inductance of the system

C is the capacitance of the system

So, the rate of rise of re-striking voltage (RRRV) is dependent upon both the inductance and capacitance of the system.

Now, the maximum value of RRRV occurs when ωnt = π/2 i.e. when t = π/2ωn

∴The maximum value of RRRV $= {V_m}{ω _n} = \frac{{{V_m}}}{{\sqrt {LC} }}$

$RRR{V_{max}} = \frac{{{V_m}}}{{\sqrt {LC} }}$

RRRV is measured in kV / μ-sec